The  Use  of  the 

Slide  Rule 


By  ALLAN  R.  CULLIMORE 

Director,  Newark  Technical  School 
NEWARK,  NEW  JERSEY 


Copyright,  1915,  1920  by 
KEUFFEL  &  ESSER  CO. 


PUBLISHED  BY 


•    KEUFFEL   &    ESSER    Co.  » 

NEW  YORK  J2Z Fulton  St.  General  Office  *ndFACtoriCs,HOBOKEN,N.  */. 

CHICAGO  ST.LOUIS  SAN  FRANCISCO  AVONTREAL, 

516-20  S.DcarbornSt.  81T  Locust  Si.  30-34  Second  St  5NotrcDameStV 

Drawin£Materials  *  Mathematical  andSurveyin^iistruments  *  Measurin&Tapes 


*       r       *      «  *      r         * 

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4-  a  .v  . 


PREFACE   TO   THE    FIRST    EDITION. 


The  need  for  a  small  book  of  this  type  arose  in  the  work  of  teaching 
the  use  of  the  slide  rule  to  engineering  and  industrial  students,  and  this  Manual 
is  a  direct  result  of  sets  of  notes  issued  to  classes  consisting  of  engineering 
students  and  men  of  more  or  less  practical  experience.  The  book  is  not  a 
treatise  in  any  sense,  its  aim  being  to  develop  the  ideas  of  the  operator  rather 
than  to  give  empirical  rules.  Those  rules  that  have  been  given  are  for  the 
purpose  of  training  the  student  in  the  formulation  of  processes,  and  it  is  not 
intended  that  they  shall  ever  be  committed  to  memory. 

The  examples  have  been  taken  largely  from  Hydraulics  and  Mechanics, 
and  while  the  actual  field  covered  by  specific  problems  is  narrow,  the  idea  has 
been  to  make  them  fundamental.  It  is  hoped  that  these  examples  will  serve 
the  purpose  of  development  better  than  the  more  specialized  problems  arising 
in  different  branches  of  engineering. 


PREFACE  TO  THE  SECOND  EDITION. 


The  second  edition  is  made  necessary  by  the  increasing  popularity  of 
the  Polyphase  type  of  rule,  a  popularity  rightly  based  on  the  combined  ef- 
ficiency and  simplicity  of  this  type.  For  students  in  engineering  and  vocational 
schools  the  Polyphase  slide  rule  is  to  be  strongly  recommended.  While  the 
general  principles  outlined  in  the  first  edition  apply  equally  to  the  Mannheim 
and  the  Polyphase  types  of  rule,  it  seems  advisable  to  make  additions  to  certain 
chapters  explaining  the  use  of  the  Polyphase  rule  when  the  use  of  this  type 
makes  for  more  efficient  calculation. 


THE  USE  OF  THE  SLIDE  TITTLE. 

I.    ACCURACY  AND  SIGNIFICANT  FIGURES. 

It  is  absolutely  necessary  for  a  proper  and  efficient  use  of  the  slide 
rule  that  the  operator  should  have  a  clear  and  definite  idea  of  the  conditions 
under  which  the  rule  may  be  used  to  advantage.  Even  among  engineers,  the 
idea  is  often  expressed  that  the  slide  rule  is  inaccurate,  because,  in  the  hands 
of  a  reasonably  expert  operator,  the  rule  will  give  results  accurate  only  to 
1/10  of  1%.  It  should  be  borne  in  mind  that  the  rule  is  inaccurate  in  exactly 
the  same  way  that  a  four  place  table  of  logarithms  is  inaccurate. 

In  a  very  large  class  of  engineering  calculations,  however,  results  which 
are  well  within  the  allowable  error  may  be  computed  by  the  slide  rule.  The 
question  most  often  arising  is  whether  the  rule  is  adapted  to  a  given  problem  or 
not.  To  answer  this,  requires  a  knowledge  of  the  proper  use  of  significant 
figures  and  a  close  inspection  of  the  data  of  the  problem,  together  with  a 
knowledge  of  the  means  employed  in  obtaining  these  data. 

By  a  significant  figure  we  mean  any  figure  which  is  significant  in  that 
it  gives  some  real  information  as  regards  the  quantity  which  is  represented. 
Thus: 

18700.1  has  six  significant  figures. 

13.7303     "  six 

0.0032       "   two 

13000        "  two  or  five  significant  figures. 

Notice  that  in  the  last  case  an  ambiguity  arises  and  that  any  one  or  all 
of  the  zeros  may  or  may  not  be  significant. 

Take  the  problem  of  finding  the  area  of  a  circle  whose  radius  is  4.67  feet, 
measured  with  a  steel  tape.  It  is  easily  seen  that  the  recorder  of  the  data 
meant  that  he  was  unable  to  state  the  distance  closer  than  1/100  of  a  foot.  In 
short,  that  he  felt  sure  that  the  distance  was  nearer  .67  of  a  foot,  than  either 
.66  or  .68.  He,  therefore,  recorded  the  result  4.67  feet,  using  three  significant 
figures  with  an  accuracy  of  one  part  in  467,  or  not  quite  1/5  of  one  per  cent.  It 
should  be  borne  in  mind  that  the  number  of  significant  figures  expresses  accuracy, 
while  the  number  of  decimal  places  may  or  may  not  do  so.  Thus  761  millimeters 
is  identical  with  .761  meters,  and  the  decimal  place  shows  nothing.     The 


4CyjJt;- 


engineer  recording  or  computing  data  should  force  himself  to  express,  by  the 
number  of  significant  figures  in  the  result,  the  accuracy  of  that  result.  The 
frequent  habit  of  carrying  results  to  a  greater  number  of  significant  figures 
than  the  data  warrant  comes  perilously  near  to  lying  with  figures;  it  certainly 
creates  a  wrong  impression  as  to  the  accuracy  of  the  result.  Certain  mathe- 
matical constants  can,  however,  be  computed  to  any  number  of  significant 
figures;  for  instance  ir  may  be  expressed  as  3.142  or  3.141592654.  On  the 
other  hand,  certain  physical  constants  are  very  uncertain,  even  in  the  third 
place.  Take,  for  example,  the  weight  of  a  cubic  foot  of  water  generally  given 
as  62.5  lbs.;  conditions  of  temperature  and  solution  may  easily  alter  the  last 
figure  of  the  three.  With  recorded  data,  however,  the  number  of  significant 
figures,  as  well  as  their  character,  gives  very  definite  information.  If  we  say 
that  light  travels  186000  miles  per  second,  we  do  not  mean  that  it  covers 
186000  miles  to  within  the  smallest  fraction  of  an  inch  in  one  second  of  time,  but 
that  the  distance  covered  is  nearer  186000  miles  than  it  is  185000  or  187000,  and 
the  accuracy  1/186  is  expressed  by  three  significant  figures.  In  this  particular 
instance,  it  will  be  noticed  that  the  three  zeros  to  the  right  of  the  six,  may  or 
may  not  be  significant  figures.  To  prevent  this  ambiguity  it  has  been  suggested 
that  results  like  the  above  be  written  186  X  (10)3  which  obviates  the  difficulty. 

If,  therefore,  the  slide  rule  will  consistently  give  results  to  within  1/10 
of  one  per  cent,  or  to  one  part  in  a  thousand,  we  have  a  right  to  use  it  where 
three  significant  figures  are  warranted  in  the  result.  The  following  rule  given 
by  Holman  should  be  rigidly  observed  in  all  cases:  "If  numbers  are  to  be  mul- 
tiplied or  divided,  a  given  percentage  error  in  one  of  them  will  produce  the  same 
percentage  error  in  the  result."  This  amounts  to  saying  that  all  problems, 
involving  data  correct  to  three  significant  figures  only,  can  be  computed 
advantageously  by  means  of  the  Slide  Rule.  The  answer  is  not  only  near 
enough,  but  is  as  accurate  as  the  data  warrant.  Of  course,  in  cases  when  the 
slide  rule  is  used  as  a  more  or  less  rough  check  on  logarithmic  or  other  calcual- 
tions,  these  questions  of  accuracy  do  not  apply.  The  consideration  of  a  very 
simple  example  will  serve  to  illustrate  the  rule  as  stated  above. 

Suppose  we  wish  to  compute  the  cubical  contents  of  a  prism  of  earth. 
Consider  that  the  horizontal  distances  have  been  measured  with  a  tape  to  the 
nearest  1/100  of  a  foot,  and  that  the  heights  have  been  measured  by  a  level  to 
the  nearest  1/10  of  a  foot,  and  the  following  dimensions  recorded: 


Multiplying: 


Length  101.13  ft.     Breadth  7.34  ft.     Depth  9.3  ft. 


101.13 
7.34 


742.2942 
9.3 

6903.33606 


Now,  if  the  answer  be  as  indicated,  we  know  the  contents  to  within  1/100000 
of  a  cubic  foot,  or  better,  with  an  accuracy  of  1/7000000  of  one  per  cent. 


—  5  — 


which  is,  of  course,  ridiculous.  Suppose  the  depth,  somewhere  between  9.25  and 
9.34:  The  correct  height  might  be  9.25,  or  9.34  but  if  tenths  alone  were  ex- 
pressed, it  would  be  recorded  in  both  cases  as  9.3.  We  see,  therefore,  that  the 
correct  result  lies  between  the  following: 


742.2942 
9.25 

6866.221350 


742.2942 
9.34 

6933.027828 


We  see,  then,  that  actually  we  know,  only,  that  the  result  surely  liesbeetwen 
6866.221350  and  6933.027828,  but  we  certainly  know  nothing  more  definite 
than  this.  If  we  express  the  answer  first  found  as  6903.33606,  we  know  nothing 
about  the  last  seven  figures.  We  are  sure  of  only  the  first  two  and  the  result 
should  have  been  written  6900.  In  the  light  of  this  we  now  perform  the  same 
multiplication  as  follows: 

101.13 
7.34 


742.0 
6900. 


giving  the  answer  6900,  which  is  as  near  the  true  value  as  we  can  know  by 
the  data  recorded. 

It  will  be  readily  seen  that  a  knowledge  of  the  proper  number 
of  significant  figures  saves  an  immense  amount  of  time  in  calculation.  This 
is  true  no  matter  what  means  are  used  in  calculating,  whether  it  be  multiplica- 
tion, logarithms  or  the  slide  rule.  The  operator  should  accustom  himself 
first  to  examine  the  data  of  a  problem  and  mentally  calculate  the  desired 
accuracy  of  the  result,  as  well  as  the  approximate  numerical  value  of  that  result. 


II.     DESCRIPTION  OF  THE  RULE. 

The  usual  type  of  engineer's  slide  rule  (K.  &  E.  Mannheim  or  Polyphase) 
is  of  wood,  faced  with  a  white  composition  upon  which  the  units  are  graduated 
in  black,  and  is  about  ten  inches  long.  Along  the  center  line  of  the  rule,  a  slide 
of  wood  moves  easily  in  a  longitudinal  groove.  Each  rule  is  provided  with  an 
indicator  or  runner  of  glass,  marked  with  a  hair  line,  which  serves  as  a  reference 
line  in  calculating.  There  are  four  distinct  scales  on  the  face  of  the  Mannheim 
and  five  on  the  Polyphase  rule,  and  for  the  sake  of  convenience  we  will  always 
call  the  scales  A,  B,  C,  D,  beginning  at  the  top  and  reading  down.  If  the 
slide  be  inverted  (that  is,  turned  upside  down  in  its  groove)  or  if  it  be  reversed 
(exposing  the  back  of  the  slide),  the  second  scale  from  the  top  will  always  be 
referred  to  asB,  the  third  as  C,  etc.  As  the  two  top  scales  are  double,  we  will 
speak  of  the  right  or  left  A,  as  the  case  may  be.  If  the  slide  be  reversed,  three 
scales  are  seen  on  the  back  of  the  slide;  a  scale  of  sines  marked  S,  a  scale  of 
tangents  marked  T,  and  a  scale  of  equal  parts  in  the  center.     Notice  on  the 


—  6  — 


back  of  the  rule  proper  a  piece  of  transparent  material  set  in  the  end  of  the 
rule  with  a  line  marked  upon  it  by  which  the  scales  can  be  read.  In  operating 
the  rule  the  slide  will  be  used  in  four  positions:  direct,  inverted,  reversed  direct, 
and  reversed  inverted.    The  terms  for  these  positions  are  explained  above. 


III.    PRINCIPLE  OF  THE  RULE. 

The  rule  is  based  on  the  principle  that  the  addition  of  the  logarithms  of 
two  numbers  gives  the  logarithm  of  the  product  of  the  two  numbers,  and  that 
the  subtraction  of  the  logarithm  of  one  number  from  the  logarithm  of  another 
number  gives  the  logarithm  of  the  quotient  obtained  by  dividing  the  second 
number  by  the  first.  If  then  we  add  a  length  on  the  rule  to  another  length,  and 
these  two  lengths  are  proportional  to  the  logarithms  of  certain  numbers,  then 
the  length  which  represents  the  sum  of  these  two  lengths  will  be  proportional 
to  the  product  of  the  two  numbers.  This  can  be  shown  quite  simply  on  the 
rule.  On  the  back  of  the  slide  we  find  in  the  center  a  scale  of  500  equal  spaces. 
Suppose  we  reverse  and  invert  the  slide,  and  bring  the  extreme  left-hand 
graduations  into  coincidence.  Set  the  runner  to  2  on  scale  D,  and  read  under 
the  runner  on  the  middle  scale  of  equal  spaces.  We  read  301.  Reading 
on  AD  in  the  same  way  we  have  602,  and  on  SD — 903.  The  distance  between 
1  and  2  is  therefore  301  units,  and  between  1  and  4  is  602  units;  adding  these 
we  would  have  903,  which,  as  we  have  seen,  corresponds  to  8.  Division  is, 
of  course,  the  reverse  of  this  process. 

In  describing  different  settings,  LA  and  RB  will  be  used  for  the  left-hand 
scale  of  A  and  the  right-hand  scale  of  B  respectively.  R  alone,  refers  to  the 
runner.  LLA  would  mean  the  left-hand  index  on  scale  A.  The  operation,  R 
to  3LA,  would  consist  in  moving  the  runner  until  the  hair  line  on  it  concided 
with  the  3  on  the  left-hand  scale  of  A.  SC  to  ALA  would  mean  placing  oC  in" 
such  a  position  that  3C  and  ALA  are  the  same  distance  from  the  end  of  the 
rule;  that  is,  both  would  be  brought  into  the  same  straight  line.  This  is  best 
done  by  placing  the  runner  so  that  the  line  on  it  is  on  the  number  on  the  fixed 
scale,  and  then  moving  the  slide  until  the  number  on  the  slide  is  under  the 
line  on  the  runner. 


IV.    DECIMAL  POINT  AND  READING  THE  SCALE. 

Success  as  an  operator  depends  upon  a  quick  and  accurate  reading 
of  the  graduations,  and  this  can  only  be  acquired  by  faithful  practice.  It 
should  be  constantly  borne  in  mind  that  there  is  no  way  to  distinguish  by  direct 
reading  alone  the  position  of  the  decimal  point.  The  reading  on  the  rule 
would  be  the  same  for  each  of  these  numbers  1751,  .1751,  17.51.  The  rule 
gives  simply  a  succession  of  figures  in  their  proper  order,  but  without  determin- 
ing the  decimal  point.  This  determination  always  must  be  made  independently 
of  the  actual  solution  of  the  problem  in  exactly  the  same  way  that  the  char- 
acteristics of  logarithms  are  independently  computed.     Suppose  the  problem 


7  — 


is  to  set  the  hair-line  on  the  runner  to  478  on  Scale  D  (R  to  478D).  We 
see  that  between  4  and  5  (care  should  be  taken  not  to  confuse  this  4  and  5  with 
1.4  and  1.5  which  occur  further  to  the  left),  there  are  20  divisions,  a  very  long 
division  marking  4.5,  45,  or  450,  whichever  we  please  to  call  it.  Between 
450  and  500  there  are  4  long  marks  and  5  short  ones;  the  first  long  one  following 
being  460,  the  next  470,  the  next  480,  etc.  Our  478  lies,  then,  between  the 
second  and  third  long  mark  to  the  right  of  450;  that  is,  between  470  and  480. 
Between  470  and  480  we  see  a  smaller  division  marking  475,  and  there  is  no 
division  between  475  and  480.  478  lies  in  this  blank  space  and  lies  3/5  or 
6/10  of  the  distance  between  the  marks  to  the  right  of  475.  In  the  same 
way,  to  set  on  1673  on  Scale  D  (R  to  1673D),  we  see  1  at  the  extreme  left  fol- 
lowed by  a  smaller  1  marking  1.1D  or  HOOD,  then  a  small  2  marking  1200D, 
etc.  1600  is  easily  found,  and  between  1600  and  1700  are  ten  divisions,  1650 
being  marked  by  a  long  one.  The  next  short  division  to  the  right  of  1650  is 
1660,  the  next  1670,  and  our  number  lies  between  1670  and  1680;  3/10  of  the 
distance  from  1670  to  1680,  to  the  right  of  1670,  lies  1673D.  This  procedure 
should  be  gone  through  over  all  parts  of  the  scale  until  the  operator  is  familiar 
with  the  different  graduations  and  can  set  on  numbers  quickly  and  accurately. 
It  will  be  seen  that  the  value  of  the  distance  between  graduations  changes  on 
different  parts  of  the  rule.  Three  significant  figures  should  be  read  on  all 
parts  of  D  and  C;  on  the  left,  four  may  be  obtained.  On  A  and  B,  two  and 
sometimes  three  are  obtainable. 

In  regard  to  the  fixing  of  the  decimal  point,  this  is,  of  course,  fixed  for 
any  given  set  of  data.  General  rules  might  be  given  for  fixing  the  decimal 
point,  but  it  seems  best  in  a  vast  majority  of  cases  to,  fix  the  decimal  point 
independently  of  the  rule.  This  is  simple  in  many  cases  and  gives  the  added 
advantage  of  a  rough  check  on  the  work.  Examples  will  be  taken  up  under 
each  class  of  computation.  It  is  sufficient  to  consider  here  a  very  simple  case: 
Find  the  square-root  of  1873.  This  evidently  has  a  square  root  lying  between 
10  and  100;  pointing  off  two  places  to  the  left  of  the  decimal  point  is,  therefore, 
correct. 

V.    EQUIVALENT  RATIOS. 

One  of  the  most  important  uses  of  the  slide  rule  is  in  converting  quantities 
from  one  kind  of  units  to  another.  Printed  on  the  back  of  the  rule  are  a  series 
of  ratios  existing  between  different  systems  of  units.  The  time  saved  by  using 
these  ratios  is  very  great,  especially  when  a  considerable  number  of  quantities 
are  changed. 

Suppose  we  wish  to  change  ounces  to  grams  and  we  have  63^,  7  %,  9}4> 
\\%,  ounces  to  be  expressed  as  grams.  We  see  on  the  back  of  the  rule  that 
6  ounces  equals  170  grams.  One  ounce  will  equal  approximately  30  grams, 
and  this  knowledge  allows  us  to  fix  the  decimal  point  easily  in  any  case.  Set 
the  runner  to  6D  and  put  170C  to  the  runner;  we  then  read  6  ounces  equals 
170  grams,  reading  ounces  on  D  and  grams  immediately  above  on  C.  Notice 
after  this  setting  is  made  that  one  ounce  equals  28.3  grams,  and  that  one 
gram  equals  .0353  ounces.  Reading  directly  over  6.5,  7.75,  9.25,  on  D,  the 
equivalent  grams  are  shown  on  C  as  follows:  184,  220,  262.     We  cannot  read 


—  8  — 


over  11.67,  as  no  scale  is  there,  so  we  shift  the  slide  to  the  left,  putting  the  right 
index  of  C  where  the  left  one  stood  previously  and  then  on  C,  by  means  of  the 
line  on  the  runner,  we  read  331  grams. 

Consider  this  example:  Find  the  intensity  of  water  pressure  in  pounds 
per  square  inch  at  depths  of  14.33, 17.84, 19.83,  9.76  feet  below  the  surface.  On 
the  back  of  the  rule  we  find  feet  of  water  :  pounds  per  square  inch  ::  60  :  26, 
or  roughly,  2  ft.  of  water  gives  a  pressure  of  one  pound  per  square  inch;  the 
fixing  of  the  decimal  point  is,  therefore,  easy.  Set  26C  to  60D  (by  means  of 
runner).  Notice  that  one  pound  per  square  inch  pressure  corresponds  to  head 
of  2.3  ft.,  and  that  1  ft.  head  gives  a  pressure  of  .434  lbs.  per  square  inch;  and 
corresponding  to  14.33  ft.,  17.84  ft.,  19.83  ft.,  and  9.76  ft.,  we  have  6.21  lbs., 
7.73  lbs.,  8.60  lbs.,  4.23  lbs.,  noticing  that  the  slide  must  be  shifted  on  the  first 
and  last  reading.  In  this  connection,  it  is  well  to  notice  that  these  settings 
for  conversion  can  be  made  on  scales  A  and  B,  and  then  no  shifting  of  the  slide 
is  necessary,  but  the  loss  in  accuracy  in  most  cases  prohibits  the  use  of  these 
scales.  Instead  of  setting  26C  to  60D  we  might  have  set  60C  to  26D  and  read 
feet  on  C  to  pounds  pressure  per  square  inch  on  D. 

Ratios  in  constant  use  in  a  special  class  of  calculations  are  sometimes 
more  conveniently  expressed  as  round  numbers.  We  see  on  the  back  of  the 
rule  the  ratio  of  the  diameter  of  a  circle  to  its  circumference  given  as  113  to  355 
instead  of  1  to  3.1416.  The  first  setting  is  somewhat  easier  and  in  all  classes 
of  work  the  change  in  ratio  is  easily  effected.  Instead  of  memorizing  a  large 
number  of  these  equivalents,  it  is  advisable  to  work  them  out  when  necessary , 
using  simply  the  few  given  on  the  back  of  the  rule.  Suppose  we  wish  the 
equivalent  of  U.  S.  gallons  of  water  expressed  in  kilograms.     We  get  the 

,      .      ,         J_,     ,     .      ,  ,,         .       U.  S.  Gallons  3      Pounds        75. 

following  from  the  back  of  the  rule:  Pounds  of  water  -    W .  ~^^  -  j; 

One  gallon  =  -5-  lbs.  One  pound  =  ==  Kilos.  Therefore,  One  gallon  = 

kilos  =  q  kilos,  or  9  gallons  =  34  kilos.   Therefore,  -=== =  — ,      which   is 

the  required  ratio.     Consider  finding  the  equivalent  of  chains  in  feet,  by  the 

865 
43 


rule,  Chains:    Meters  ::  43  :  865;  Yards  :  Meters  ::  82  :  75;  1  Chain  = 


Meters;  1  Meter 
865  X  82  X  3 


82 
75 


Yards;   1   Yard    =   3   Feet.     Therefore,   1   Chain  = 


Feet;  Chains  :  Feet  ::  215  :  14200,  or  as  1  is  to  66,  which  is  the 
4o  X  75 

correct  answer.  These  examples  will  serve  to  show  how  any  desired  equivalent 
may  be  obtained.  They  may  easily  be  solved  by  following  the  instructions 
which  will  be  given  for  continued  multiplication  and  division. 

The  slide  rule  also  furnishes  a  ready  means  of  converting  inches  into 
decimals  of  a  foot,  and  decimals  of  an  inch  into  any  desired  fractional  part 
thereof;  it  facilitates  inverse  operations.  To  reduce  the  number  of  inches  and 
fractions,  given  to  the  nearest  \  in.,  to  decimals  of  a  foot,  setlC  to  12D  and  over 


9  — 


inches  on  D,  find  feet  and  tenths  on  C.  For  example  to  find  the  decimal  of  a  foot 
corresponding  to  3  M  inches,  read  above  3.25D  and  get  .2708  on  C.  This  setting 
can  be  conveniently  used  when  the  distances  are  given  to  the  nearest  }4  or 
perhaps  \i  in.  We  must,  of  course,  remember  the  decimals  of  an  inch  correspond- 
ing to  each  eighth  of  an  inch  as  follows:  .125,  .250,  .375,  .500,  .625,  .750,  .875. 
The  reduction  to  decimals  from  sixteenths  is  simple.  Set  \C  to  16D  and  over 
the  number  of  sixteenths  on  D  find  the  decimal  on  C.  As  can  readily  be  seen, 
this  method  is  applicable  to  any  fractional  part,  such  as  a  sixty-fourth. 

Take  the  following  example:  Express  7  9/64  in.  as  a  decimal  of  a  foot. 
Put  \C  to  64D,  using  the  left-hand  index  on  C.  Reading  on  C  over  9  on  D, 
we  find  .141.  Set  1C  to  12D  and  read  .595  on  C,  over  7.141  on  D.  These 
examples  are  really  special  cases  of  a  general  rule.  To  reduce  fractions  to  deci- 
mals, set  numerator  on  C  to  denominator  on  D.  Read  on  C  over  ID  to  find 
equivalent  decimal,  and  to  reduce  decimals  to  fractions,  set  decimal  on  C  to  ID, 
and  corresponding  numerators  will  be  found  over  their  denominators. 

.31 

Example:    Find  the  decimal  corresponding  to  — .    Set   31C   to   47D, 

47 
60      198 

and  over  ID  read  .66  either  —  or .    We  are  enabled  to  find  reciprocals  in 

91      300 
much  the  same  way.     Example:    Find  the  reciprocal  of  3.17.    Set  3.17C  over 
ID,  and  read  on  D  under  \C.    Answer  is  .316. 

On  the  back  of  the  rule  will  be  noticed  gauge  points  for  different  metals, 
with  the  weight  of  one  cubic  foot  of  the  metal  given.  In  using  these  ratios,  it 
is  convenient  to  reduce  all  weights  to  equivalent  weight  of  iron,  and  then 
reduce  by  means  of  ratios  given.  Suppose  we  wish  to  make  a  table  of  weights 
of  brass  plates  12  in.  on  a  side,  varying  by  32nds  of  an  inch  in  thickness. 

480 
Dividing =  40,  which  is  the  weight  of  1  square  foot  of  iron,  1  in.  thick;  set 

12 
32C  to  40D,  and  multiply  in  each  case  by  1.09,  the  gauge  point  for  brass.  ]  The 
best  method  for  this  setting  is  toset32C  to  40D,  R  to  1.09C.    Answer  1.36D. 
Then  Rto2C,  ICto  R,R  to  1.09C.    Answer  2.72  D,  etc.  ^  =  1-36;  TV  =  2.72; 
T35  =  8.16,  etc.    Or  we  might  have  done  as  follows: 

525 

=  43.7.    Set  32C  to  43.7D  and  we  have  as  before  1.36,  2.72,  8.16. 

12 
This  same  idea  is  applicable  to  any  section  of  bar,  plate  or  structural  shape. 
(See  page  10). 


VI.  SQUARES  AND  SQUARE  ROOTS. 


The  numbers  on  A  are  the  squares  of  those  on  D.  We  have  then  a 
table  of  squares  and  square  roots  ready  at  hand.  To  find  the  square  of  any 
number  set  the  runner  to  the  number  on  D  and  read  the  answer  on  A .  Example : 
to  find  the  square  of  4.17,  set  the  runner  to  417D  and  find  under  runner  on  A 
17.4.  The  position  of  the  decimal  point  is  very  evident  as  shown.  Finding 
the  square  root  is  the  reverse  of  this  process.    Set  the  runner  to  any  number 


—  10  — 

IIHl 

^■(^^k^'^^P^'^l      KEUFFEL  &   ESSER  CO..  NEW  YORK      lv,*S3l 

^^>Q^^i^^^^SSrlllSjrr^rr                     '                    '            ***^ 

on  A  and  under  the  runner  find  the  square  root  of  the  given  number.  Care  must 
be  taken  to  choose  the  proper  scale  on  A.  In  finding  the  square  root  of  5,  if 
we  set  the  runner  on  5LA,  we  find,  under  the  runner  on  D,  2.24,  and  if  we  use 
5RA,  we  find  7.07.  It  is  evident  that  2.24  is  the  square  root  of  5,  and  7.07  the 
square  root  of  50.  We  can  use  this  rule :  If  a  number  has  an  uneven  number  of 
digits  to  the  left  of  the  decimal  point,  use  the  left-hand  scale;  if  an  even  number, 
the  right-hand  scale  on  A.  Thus  we  see  that  under  the  left-hand  5  we  would  find 
the  square  roots  of  5,  500,  50000,  which  would  be  2.24,  22.4,  224.  It  is  perhaps 
better  in  most  cases  to  estimate  the  result  mentally  and  set  on  the  scale  so  as 
to  approximate  this  result.  Estimating  that  the  square  root  of  5  would  be  2, 
we  can  quickly  see  that  the  left-hand  scale  must  be  used  to  obtain  an  answer 
in  the  neighborhood  of  2. 


We  have 


Suppose  we  wish  to  find  th 
525  X  22  X  d2 


weight  in  lbs.  per  lineal  foot  of  brass  rods. 
51  Xrf2 


lbs.  which  is  equal  (almost)  to 


18 


We  can  solve 


144  X  7  X  4 

then  very  easily  in  the  following  manner:  Set  1SB  to  51A,  read  answer  on  A 
over  diameter  in  inches  on  C.  In  setting  1SB  to  51A,  the  operation  has  been 
performed  with  the  exception  of  multiplying  by  the  variable  d2.  Now  d2  on 
B  corresponds  to  d  on  C,  so  multiplying  by  d  on  C  is  equivalent  to  using  d2  on 
B,  hence  the  setting  as  given. 


VII.     MULTIPLICATION  AND  DIVISION. 


Notice  that  if  we  place  1C  to  2D,  every  number  on  D  is  twice  the  number 
immediately  above  it  on  C;  that  is,  we  have  formed  a  series  of  proportions, 
using  the  ratio  1  to  2.  All  numbers  on  C  have  this  ratio  to  the  numbers  im- 
mediately below  on  D.  Suppose  that  we  wished  to  solve  the  proportion  3.8  : 
2.18  ::  x  :  4.53.  Set  3.8C  to  2.18D  (always  use  the  runner  to  make  this  class 
of  setting)  and  over  4.53D  find  7.9C.  In  making  this  setting,  place  the  runner 
first  on  2.18D  and  then  slide  C  along  until  3.8  comes  under  the  runner;  then  slide 
the  runner  to  4.53D  and  read  7.9  on  C,  which  is  the  answer.  This  gives  us  a 
method  of  conveniently  performing  the  operations  of  multiplication  and  division. 
These  operations  can  be  performed,  using  the  two  top  or  the  two  bottom  scales, 
as  determined  by  the  percentage  accuracy  required  in  the  result.  To  multiply 
on  the  bottom  scale,  set  \C  at  a  factor  on  D;  under  the  other  factor  on  C 
read  the  answer  on  D.  Setting  is  made  as  follows:  Slide  C  until  1  is  over  a 
factor  on  D,  slide  the  runner  to  the  other  factor  on  C,  and  read  the  answer  on  D 
under  the  runner.  Example:  3.26X2.83.  The  answer  is  estimated  as 
approximately  9.0.  Set  1C  to  3.26D  and  under  2.83C  read  9.22  on  D,  which  is 
the  answer.  Example:  3.91  X  7.33.  The  answer  is  estimated  to  be  about 
28.  Set  1C  to  3.9LD,  under  7.33C-find  28.7,  answer.  In  this  second  example,  it 
will  be  noticed  that  we  were  forced  to  use  the  right-hand  \C  instead  of  the  index 
to  the  left.  A  very  considerable  amount  of  time  will  be  saved  if  the  operator 
decides,  before  trying,  which  index  he  will  have  to  use.  The  following  example 
will  serve  to  illustrate:     13.14  X  16.93.     The  answer  will  lie  between  100  and 


— 11  — 


300.  As  to  using  the  right-hand  or  left-hand  scale,  we  suggest  the  following 
rule:  When  the  first  digits  of  the  factors  are  such  that  their  product  gives  a 
number  greater  than  ten,  use  the  right  index  on  C;  otherwise  use  the  left. 
Take  as  examples  the  following: 

2.13  X  3.33,  2X3=6.      Use  the  left  index. 
7.23  X  4.71,  7  X  4  =  28.        "     "    right  " 
.131  X  4.6,     1X4=4.  "    "    left      " 

If  this  very  simple  rule  be  kept  in  mind  much  time  will  be  saved.  It  will 
be  a  little  difficult  in  certain  cases,  like  the  following,  to  make  a  decision: 
3.13  X  3.31.  By  the  rule,  the  left-hand  scale  should  be  used,  considering  only 
the  first  digit  of  each  of  the  factors;  a  closer  inspection  will  show,  however,  that 
this  is  not  correct,  and  that  the  right  index  should  have  been  used.  The 
operator  should  not  try  to  memorize  these  rules,  but  he  should  appreciate 
their  significance.  The  one  just  given,  if  appreciated,  will  save  an  immense 
amount  of  time.  If  the  process  of  multiplication  be  carried  out  between  scales 
A  and  B,  and  if  we  always  use  the  middle  index  on  B,  the  second  factor  will 
never  fall  outside  the  scale  of  A. 

Division  is  the  reverse  of  the  process  of  multiplication.  Set  the  divisor 
on  C  to  the  dividend  on  D,  read  the  answer  on  D  under  1C  (The  runner  should 
be  used  in  setting  the  divisor  on  the  dividend.)  Example:  Divide  313  by  8.9; 
the  answer  is  estimated  to  lie  between  30  and  40.  Set  8.9C  to  313D  and  read 
under  \C  the  answer  35.2.  Notice  that  in  the  case  of  division  only  one  index 
can  appear  on  the  scale,  so  that  no  ambiguity  can  possibly  arise. 

Attention  should  be  called  at  this  point  to  a  very  simple  setting  for 
finding  the  areas  of  circles.  Suppose  that  we  use  scales  A  and  B.  Required 
the  area  of  a  circle  whose  diameter  is  11.6  inches.     We  use  the  formula  A  = 

=  .7854  d2.     (Notice  that  .7854  is  marked  on  scales  RA  and  RB  by  a  longer 

4 
line  in  the  same  way  that  -n-  is  marked  on  the  same  scales.)  We  might  solve 
this  by  simple  division,  but  the  following  will  be  found  easier.  R  to  11. 6D  (see 
the  square  under  the  runner  on  A),  set  IB  to  R,  and  over  .78545  read  the  answer 
on  A.  It  is  seen  to  be  105.7.  Or  simpler  still  is  the  following:  Reading  on  the 
back  of  the  rule:  Diameter  circle  :  side  of  equal  square  ::  79  :  70.  Set 
79C  to  70D,  R  to  11.6C,  and  under  runner  read  the  answer  on  A,  105.7.  The 
advantage  of  this  last  method  is  apparent,  for  after  the  first  setting  of  79C  to 
70D  the  area  of  any  circle  may  be  found  by  one  setting  of  the  runner.  The 
slide  rule  is  primarily  a  time-saving  device  and  the  idea  always  should  be  to 
choose  the  setting  giving  the  quickest  solution.  In  this  special  case,  the  last 
setting  will  always  prove  the  most  efficient  when  the  number  of  areas  is  greater 
than  two  or  three. 

In  engineering  work,  formulae  very  frequently  take  the  form  of  a  fraction 
in  which  the  denominator  and  the  numerator  are  made  up  of  several  factors. 

142  X  15  X  15  X  45000 

Consider  the  solution  of  the  following  example: ;  first 

4  X  16  X  16 

determine  the  decimal  point  mentally.      The  two   16s   cancel  the   two  15s 


—  12  — 


approximately,  4  into  142  goes  35  times,  45  X  35  =  1000,  add  three  ciphers  and 
the  answer  will  be  approximately  1600000.  This  process  of  determining  the 
decimal  point  should  always  be  followed,  no  matter  what  the  problem. 

It  is  evident  that  we  might  multiply  all  the  factors  of  the  numerator  first 
and  divide  the  result  by  each  of  the  factors  of  the  denominator  in  turn.  Or 
we  might  alternately  multiply  and  divide.  The  second  method  is  much  simpler 
and  shorter,  and  an  actual  comparison  of  the  two  methods  in  this  problem  will 
show  the  gain  in  speed  of  the  second  method  over  the  first.  The  first  method : 
Set  1C  to  1.42D,  R  to  15C,  1C  to  R,  R  to  15C,  1C  to  R,  R  to  45C  (note  here  a 
change  of  index),  AC  to  R,  R  to  1C,  16C  to  R,  R  to  1C,  16C  to  R,  read  under  \C 
the  answer  1404000  on  D.  The  second  method:  Set  16C  to  142D,  R  to  15C; 
16C  to  R,  R  to  15C,  AC  to  R;  read  under  the  45C  the  answer  1404000  on  D.  It 
will  be  seen  that  the  work  of  the  first  method  is  almost  double  that  of  the 
second.  The  success  of  this  method  lies  in  the  correct  choosing  of  divisors. 
Suppose  that  the  example  had  been  solved  as  follows:  AC  to  142D,  R  to  15C, 
16C  to  R,  R  to  15C,  16C  to  R,  R  to  1C,  1C  to  R,  read  under  45C  1404000  on  D. 
This  process  is  evidently  more  laborious.     It  is  not  always  possible  to  arrange 

45X130X101 

divisors  so  as  to  prevent  the  shifting  of  indices.  Take  for  example, :- 

72X84 

mentally,  45  into  84  twice,  2  into  130,  65  times,  72  into  6500  about  90,  which  is 
the  approximate  answer.  Probably  the  best  solution  is  1C  to  45D,  R  to  13C, 
72C  to  R,  R  to  1C,  84C  to  R,  R  to  1C,  1C  to  R,  read  the  answer  on  D  under 
101C  as  97.7 ;  another  solution  is  1C  to  45D,  R  to  13C,  72C  to  R,  R  to  1C,  1C  to  R, 
R  to  101C,  84C  to  R,  read  under  1C  the  answer  97.7  on  D. 


The  solution  of  the  following  types  is  almost  as  simple  as  that  of  simple 


division: 


b'    b2'     b2'    ^b"' 


VjL      1-1 
b  '   \b 


Int—,  R  to  aD,  bB  to  R.  Answer  on  A  over  LB. 

In  -f^>  R  to  aD,  bC  to  R.  Answer  on  A  over  LB. 
b2 

In  ■§;,  R  to  aA,  bC  to  R.  Answer  on  A  over  LB. 


'^ 


In  — =•»  R  to  aD,  bB  to  R.     Answer  on  D  under  1C 

In  y  a     R  to  aA,  bB  to  R.    Answer  on  D  under  1C. 
b ' 


sl 


In    I—,   R  to  aA,  bB  to  R.     Answer  on  D  under  1C 


The  thorough  mastery  of  the  principles  involved  in  these  settings  will 
suggest  others  as  simple. 


—  13  — 


VIII.     THE  USE  OF  THE  RULE  WITH  INVERTED  SLIDE. 


Invert  the  slide  so  that  the  former  scale  C  is  adjacent  to  A;  scale  C  now 
becomes  B,  and  B  becomes  C,  A  and  D  remaining  the  same.  If  we  make  the 
indices  of  the  slide  to  coincide  with  the  indices  of  the  stationary  scale,  we  have, 
reading  from  scale  D  to  scale  B,  a  table  of  reciprocals.  Set  R  to  AD  and  read 
scaleU  under  R,  and  we  see  .25,  the  reciprocal  of  4.  The  position  of  the  decimal 
point  is  easily  determined  by  inspection.  Find  the  reciprocal  of  31.5.  100/32 
is  about  3,  so  the  approximate  answer  will  be  .03.  Set  R  to  31.5D,  and  reading 
under  the  runner  onB,  we  find  318,  or  as  we  know  it  to  be,  .0318.  This  gives 
ready  means  for  converting  fractions  of  feet  or  inches  into  decimals,  provided 
the  numerator  is  unity.  Set  R  to  64D,  reading  onB  under  the  runner  .01562. 
Or  1/64  of  an  inch  =  .01562  in. 

Multiplication  and  division  can  readily  be  performed  on  the  rule  with 
the  slide  inverted,  the  process  being  in  each  case  the  reverse  of  the  process 
with  the  slide  direct.  Invert  the  slide  and  set  R  to  AD,  and  2B  to  R  and  under 
LB  find  8  on  D.  In  general,  then,  to  multiply,  set  the  runner  to  one  factor  on 
D  and  bring  the  other  factor  onB  under  the  runner,  and  look  under  IB  on  D  for 
the  answer.  With  slide  inverted,  set  R  to  AD,  and  LB  to  R,  and  under  2B 
find  2  on  D;  to  divide,  therefore,  set  the  runner  to  dividend  on  D,  set  IB  to  the 
runner  and  find  the  answer  on  D  under  the  divisor.  It  will  be  easily  seen  that 
multiplication  with  the  slide  inverted  is  simpler  than  the  same  process  with  the 
slide  direct,  and  all  problems  involving  only  the  product  of  two  or  more  quantities 
should  be  solved  with  the  slide  inverted.  In  multiplying  with  the  slide  direct, 
we  found  that  the  question  arose  as  to  which  of  the  indices  to  use.  This 
question  cannot  arise  with  the  slide  inverted.  The  only  disadvantage  con- 
nected with  the  inverted  slide  is  that  scales  B  and  D  are  separated,  but  this  is 
more  than  offset  by  the  absence  of  ambiguity  in  choosing  indices,  and  by  the 
fact  that  both  right  and  left  indices  on  the  slide  coincide.  As  an  example  take: 
137  X  16.3  X  8.13;  by  inspection  the  answer  will  be  near  21000.  Solve  as 
follows  with  the  slide  inverted:'  R  to  137D,  163B  to  R,  R  to  LB,  8135  to  R,  and 
under  LB  find  the  result  on  D  as  18150. 

With  the  slide  direct  we  would  have  the  following:  \C  to  137D,  R  to 
163C,  \C  to  R,  note  here  a  change  of  index,  R  to  813C,  and  read  result  on  D  under 
R,  as  18150.  The  advantage  of  the  first  method  is  obvious.  Both  multiplica- 
tion and  division  with  the  slide  inverted  are  necessary  to  the  speedy  solution 
of  some  of  the  more  complicated  formulae  which  will  be  shown  later,  and  they 
should,  therefore,  be  mastered. 

The  inverted  scale  is  also  useful  in  cases  of  equal  products  of  the  general 
type  ax  =  be,  where  x  is  an  unknown  quantity.  With  the  slide  inverted,  set 
LB  to  8Z>  and,  with  the  runner  set  at  random,  the  products  of  the  numbers  under 
the  runner  on  B  and  D  will  always  equal  8.  Consider  the  following  example : 
A  16  in.  pulley  makes  137  R.  P.  M.;  required  the  diameters  of  pulleys  driven 
by  the  specified  pulley  so  as  to  give  speeds  of  150-175-200-250-300  R.  P.  M. 
With  inverted  slide,  set  16B  to  137D,  and  read  the  required  diameters  onB  over 
the  speeds  on  D.  We  see  corresponding  to  a  speed  of  150  a  diameter  of  14.6 
inches,  to  175 — 12.5,  etc.  This  problem  depends  upon  the  fact  that  the  speed 
of  a  pulley  (belted  or  meshing)  in  revolutions  per  minute  multiplied  by  the 


—  14  — 


diameter  of  the  pulley  is  a  constant  for  a  constant  rim  speed.  We  have  found 
this  constant  and  have  divided  it  by  the  speeds  as  given  and  found  the  respective 
diameters.  Notice  that  the  indices  must  be  interchanged  between  speeds  of 
200  and  250  R.  P.  M. 

Consider  the  reduction  of  inches  and  decimals  to  inches  and  sixty-fourths 
with  the  scale  inverted.  Required  to  express  14.6  in.  as  inches  and  sixty- 
fourths.  Set  R  to  6D,  64B  to  R,  the  result  will  be  found  on  D  under  LB,  and  38 
will  be  seen  to  be  the  nearest  whole  number.  So  the  answer  is  14-38/64  inches. 
Now  set  R  to  5D,  64JB  to  R,  and  under  LB  find  32,  as  it  should  be.  This  setting 
is  rather  inconvenient  when  a  great  many  decimals  are  to  be  changed  and  the 
following  is  preferable.  Set  R  to  64D,  6B  to  R,  under  LB  find  38  as  before.  Set 
f>B  to  R,  find  32,  etc.  It  will  be  seen  that  we  really  divide  the  decimal  by  the 
reciprocal  of  64;  this  second  method  will  prove  most  efficient  when  more  than 
two  decimals  are  to  be  converted.  As  a  further  example  of  equal  products, 
consider  this  example:  Two  vertical  pipes  are  connected  by  a  tube.  The 
cross  sectional  area  of  pipe  A  is  3.42  square  inches,  and  the  cross  sectional 
area  of  B  is  13.61  square  inches.  If  these  pipes  contain  water,  and  the  surface 
in  A  is  depressed  8.19  inch,  what  will  be  the  increase  in  height  in  B?  We  have 
a  case  of  equal  products  as  in  the  previous  example.     We  might  solve  by  con- 

3.42  X  8.19 

sidering  x  =  and  solving  with  the  slide  direct.     The  answer  is 

13.61 
approximately  2  by  inspection.  With  slide  direct,  set  1361C  to  342D,  R  to  1C, 
\C  to  R,  and  under  8.19C  find  2.06  on  D.  Notice  in  this  setting  that  the  runner 
must  be  used  to  make  the  first  and  last  setting  and  that  in  addition  the  indices 
must  be  shifted,  making  in  all  five  operations.  With  slide  inverted,  set  819B 
to  342D,  R  to  LB,  shift  indices  and  under  136LB,  find  2.06.  If  conditions  of 
accuracy  permit,  we  may  use  scales  A  and  C  when  the  slide  is  inverted;  this  is, 
of  course,  shorter  in  that  the  indices  are  not  shifted.  Set  819C  to  342A,  R  to 
1361C,  read  answer  on  A  under  R,  as  2.06.  We  might  have  set  R  to  136LA  and 
read  answer  on  C,  giving  the  same  result,  2.06. 

THE  POLYPHASE  SLIDE  RULE. 

Multiplication,  division  and  continued  multiplication  and  division  are 
very  much  simplified  by  the  use  of  the  Polyphase  rule.  In  general  50%  saving 
in  time  can  be  made  if  the  relations  of  C,  CI  and  D  are  understood.  The  red 
scale  CI  obviates  the  necessity  of  inverting  the  slide  and  all  problems  calling 
for  inverted  slide  can  be  readily  solved  with  slide  direct  by  using  scale  CI. 
The  study  of  a  typical  problem  in  detail  will  serve  to  illustrate  the  method  and 
advantage  of  using  CI.  12.8  X  6.14  X  183.2  X  62.5.  Solving  :  First  using 
C  and  D,  we  have,  R  to  12.8D,  L1C  to  R,  R  to  6.14C,  R1C  to  R,  R  to  183.2C, 
L1C  to  R,  R  to  62. 5C.     Answer  on  D  under  R  =  90000. 

This  requires  seven  settings.  Solving  :  Now  using  C,  CI.  and  D  we 
have  R  to  12.8D,  6.14CL  to  R,  R  to  1.83C,  62.5CJ  to  R.  Answer  on  D 
under  \C  =  90000.  This  requires  four  settings.  With  no  possible  ambiguity 
as  to  right  or  left  index.     Notice  R  to  12.8D,  6.14CL  to  R  gives  a  multiplication 


15  — 


answer  under  \C  on  D,  to  multiply  this  by  1.83  we  need  only  to  put  runner  to 
1.83,  answer  is  then  under  R  on  D.  Setting  62. 5CI  to  R  multiplied  by  62.5 
and  the  answer  is  on  D  under  1C.  In  general  for  simple  multiplication  it  is 
advantageous  to  set  two  factors  together  and  read  the  answer,  rather  than  to 
set  the  index  on  one  and  the  runner  to  the  other  on  the  slide.  It  is,  of  course, 
very  simple  to  remember  that  the  processes  of  multiplication  and  division  be- 
tween Ci"  and  D  are  just  the  reverse  of  those  between  C  and  D.  In  this  type 
of  rule  all  the  advantages  of  both  direct  and  inverted  slide  are  available  for 
any  particular  problem. 

326 

Run  through  the  following:     R  to  32.6D,  18.44C  to 

18.44  X  1.68  X  47 

R,  notice  here  why  CI.  can  be  used  to  advantage,  and  the  next  setting  is  R  to 
47C.1  and  16.8C  to  R.  Answer  on  D,  under  1C  =  .224.  The  student  should 
not  leave  this  subject  until  he  sees  clearly  the  interrelation  between  C  D  and  CI. 

It  frequently  happens  that  we  wish  to  successively  divide  a  number  by 
various  other  numbers.  Suppose  we  time  85  revolutions  of  a  current  meter  in 
47.3,  48.1,  46.4  seconds,  respectively,  and  wish  to  reduce  to  revolutions  per 
second.  Scale  D  with  CI  should  be  used.  Set  1C  to  85D  and  read  under 
47.3,  48.1,  46.4  and  we  have  1.80, 1.77, 1.83  revolutions  per  second,  respectively. 

IX.    THE  FORM  XY  =  B  AND  X2Y  =  B. 

Let  us  now  consider  problems  of  the  general  type  ab2  =  x;  that  is,  to 
find  the  product  of  two  factors,  one  of  which  is  the  square  of  a  given  number. 
Find  (3)  2  X  9.  Set  1C  to  3D,  read  answer  on  A  over  9B,  which  is  81.  It  is 
evident  that  we  first  square  3  and  multiply  the  answer  by  9  on  scales  A  and  B. 
The  reverse  of  this  problem  occurs  in  the  designing  of  beams,  the  problem  of 
design  being  to  determine  the  dimensions  of  a  section  of  a  beam  to  withstand 
a  given  moment.  As  this  problem  is  typical  of  the  use  of  the  slide  rule  in  a  wide 
variety  of  problems,  it  will  be  taken  up  more  in  detail  than  the  problem  itself 
might  seem  to  warrant.  Consider  the  design  of  a  rectangular  wooden  beam 
to  withstand  a  moment  of  50000  in.  pounds.  We  have  from  statics  M  =  1/6 
fbh2,  where/  =  allowable  stress  in  cross  bending  in  pounds  per  square  inch,  b  the 
width,  and  h  the  depth  of  the  beam  in  inches.  Let  1200  lbs.  equal  /,  and  we 
have  50000  =  200  bh2  or  250  =  bh2,  and  we  wish  to  determine  b  and  h  so  that 
their  product  shall  be  250  or  very  slightly  in  excess  of  that  figure.  It  is,  of 
course,  the  exact  reverse  of  the  problem  above.  The  problem  as  given  is  evi- 
dently indeterminate,  as  for  every  valve  of  b  assumed  we  can  find  a  corresponding 
value  for  h.  There  are,  as  a  matter  of  fact,  certain  limitations.  We  desire  a 
beam  of  minimum  material,  and  as  a  commercial  size  must  be  used,  we  desire 
a  beam  whose  theoretically  desired  section  shall  vary  as  little  as  possible  from 
the  commercial  sizes.  For  the  purposes  of  this  problem  let  us  say  we  can 
obtain  beams  2  X  6,  2  X  8,  2  X10,  4  X6,  4  X  8,  4  X  10,  6  X  6,  6  X  8,  6  X  10. 

Set  R  to  250A  (the  left-hand  250).  The  reason  for  this  choice  will  be 
pointed  out  later.  Slide  left  LB  to  the  runner.  We  then  see  that  (15.8)2  X  1  = 
250,  or  a  beam  1  in.  in  width  would  require  a  depth  of  15.8  in.  Put  2B  under 
the  runner,  h  then  would  have  to  be  greater  than  11  in.,  but  2  X  10  is  our 


—  16 


deepest  commercial  size.  Put  4B  under  the  runner  and  we  see  by  the  rule  that 
h  =  7.9,  that,  therefore,  a  4  X  8  would  be  sufficient  with  a  waste  of  .1  of  an  inch 
in  height  X  4  in.  in  width  or  .4  square  ins.  Put  QB  to  runner,  /i=6.47,  or  we 
would  require  a  6  in.  X  8  in.,  as  the  smallest  commercial  size  available  with 
a  width  of  6  in.  We  have  seen  that  a  4  X  8  is  sufficient;  the  6X8  then  is 
much  too  large.  Notice  that  for  each  assumed  width  the  rule  shows  the  waste 
between  the  best  theoretic  section  and  the  commercial  section  of  the  same 
width.  Now  as  to  the  choice  of  the  right  or  left-hand  scale  on  A.  As  a  matter 
of  fact,  either  may  be  used  if  the  proper  scale  on  B  is  used.  Suppose  we  had 
used  LA  with  RB,  for  b  =  4  we  would  have  found  h  =  25.  Obviously  (25)2 
X  4  is  not  equal  to  250,  .but  (2.5)2  X  40  =  250  and  such  a  choice  would  theo- 
retically fulfill  the  conditions  of  our  problem.  No  ambiguity  will  arise  if  we 
roughly  check  our  choice  once  for  each  problem. 

Let  us  consider  one  more  example:  1225  =  bh2.  Set  R  to  1225ZA, 
set  8LB  to  R;  we  find  h  equals  approximately  40.  Now  (40)2  X  8  does  not 
equal  1225.  So  use  8RB  and  we  find  h  =  approx.  12.5  and  (12.5)2X  8  = 
approx.  1225.  Rules  might  be  given  for  this  class  of  calculation,  but  the  above 
method  is  simpler  and  more  satisfactory.  Use  either  scale  on  A  and  determine 
by  trial  the  scale  to  be  used  on  B. 

Consider  another  problem  of  the  same  type  but  simpler  in  'principle. 
We  desire  to  determine  the  number,  width,  and  thickness  of  steel  plates  to  be 
used  in  the  flange  of  a  plate  girder.  The  plates  must  have  a  gross  sectional 
area  of  33.6  square  inches.  They  may  vary  in  width  from  12  to  16  by  half 
inches  (that  is,  we  may  use  12,  12^2,  13,  13  H,  etc.)  and  the  plates  may  vary 
in  thickness  from  %  in.  to  %  in.  by  16ths  of  an  inch  (that  is,  we  may  use 
H,  7/i6>  M>  etc.,  up  to  %).  Required  the  number  of  plates,  their  width  and 
thickness,  so  that  their  total  sectional  area  will  be  as  little  in  excess  of  33.6  in. 
as  possible.  The  plates  must,  of  course,  be  all  the  same  width,  but  each  plate 
may  vary  in  thickness  between  the  limits  specified;  further  it  is  desired  to  have 
the  plates  of  as  nearly  uniform  thickness  as  possible.  Set  R  to  33.6D,  set  12C 
to  R.    Read  under  16C,  44.8  on  D.    This  means  that  with  a  width  of  12  in.  the 

44.8 

plates  must  total ,  but  we  consider  only  full  16ths  according  to  the  con- 

16 
ditions  of  the  problem,  so  45/16  are  required.     Tabulating  waste  of  each  width 
from  12  to  16,  sliding  the  runner  to  12^,  13,  etc.,  to  determine  16ths  required, 
we  have : 

12         12J4     13         13^     14         14%     15         15}^     16 
.25  0         .7         .2         .6  9  2  3  4 


So  12  Yl  shows  least  waste,  and  putting  1232  under  runner  we  have  43/16  as 
total  thickness.  Now  two  whole  numbers  the  product  of  which  is  nearest  43 
are  7  and  6,  so  use  five  plates  Vie  X  12  Y%  and  one  plate  Yz  X  12.5. 

5  X  12.5X  7 
Checking:     - 


16 


12.5 
=  27.3  and =  6.25,  and 


27.3 
6.25 


33.55  chk. 


—  17  — 


THE  POLYPHASE  SLIDE  RULE. 

With  the  Polyphase  rule  cases  of  xy  =  B  are  sometimes  simpler  of  solution 
using  CI.  in  conjunction  with  D.  Consider  the  problem  as  given  above. 
Set  16CI  to  33.6  D.  Run  to  12CI,  read  under  R  on  D,  44.8,  which  is  the 
number  of  16ths  required  in  the  total  thickness  of  plates  with  a  wastage  shown. 
Shift  R  to  12.5CI  and  we  see  43,  with  no  wastage.  R  to  13  CI  under  R  41.3 
with  .7  wastage,  so  12  %  width  plates  should  be  used. 

X.     CUBES. 

Cubing  a  number  (a)  is,  of  course,  the  same  as  multiplying  a  by  b2  when 
(a)  and  (b)  are  equal.  To  cube  2.17,  set  R  to  2.17D.  Set  IB  to  R  and  over 
2.17J5  read  answer  10.2  on  A.  It  is  frequently  of  advantage  to  cube  a  number 
with  the  slide  inverted.  This  setting  is  in  reality  simpler  than  with  the  slide 
direct  and  should  always  be  used  in  connection  with  problems  in  which  for 
any  reason  the  slide  is  in  the  inverted  position.  To  cube  2.17,  set  R  to  2.17P, 
set  2.17C  to  R,  read  over  IB  the  answer  10.2  on  A.  Notice  that  this  operation 
consists  in  squaring  the  number  in  the  regular  way,  and  then  multiplying  the 
result  by  the  number  itself;  using  scales  A  and  C.  It  is  interesting  to  notice  that 
if  we  set  R  to  2.Y1D,  2.17B  to  R  and  read  over  2.17C  on  A,  we  have  the  answer 
on  A,  10,2. 

The  application  of  these  last  settings  is  clearly  seen  when  we  wish 
to  raise  a  number  to  the  3/2  power.  This  can  be  performed  very  simply  with 
the  slide  direct,  but  the  solution  with  slide  inverted  is  in  many  ways  more 
satisfactory.  Let  us  consider  the  first  problem  with  the  slide  direct.  To  raise 
3.163  to  the  3/2  power.  Set  \C  to  3.163D  and  read  the  answer  on  D  under 
3.163B,  which  is  5.63.  Notice  if  we  had  used  the  right-hand  index  instead  of 
the  left  we  would  have  read  1780.  Suppose  we  perform  the  same  operation 
with  the  slide  inverted.  Set  R  to  3.163D  and  3.163C  to  R,  read  answer  on  B 
under  \A,  5.63,  or  on  D  under  1C.  We  will  notice  that  two  readings  under  (1) 
are  possible;  one  reading  being  5.63  and  the  other  1780.  There  will,  of  course, 
never  be  any  doubt  as  to  which  of  these  two  values  is  correct  in  a  particular 
example,  if  the  operator  forms  the  habit  of  estimating  the  result.  In  this  case, 
(3.16)3  is  approximately  30,  and  V30  ^es  between  5  and  6.  As  to  the  value 
1780,  it  is,  of  course,  (31.63)  |;  the  two  values  arising  in  finding  the  square  root 
of  31.63  and  3.163  cubed.  In  making  this  setting,  notice  that  under  the  runner 
on  scale  C  is  the  number  itself,  on  A  is  the  square  of  3.163,  on  B  is  \/(3.163)3. 
Under  one  of  the  C  indices  on  D  read  v/(3.163)3an(i  under  the  other  v  3.163. 

THE  POLYPHASE  SLIDE  RULE. 


Below  D  on  the  edge  of  the  polyphase  will  be  found  a  scale  of  cubes.  Set 
R  on  any  number  on  D  and  read  the  cube  of  that  number  on  the  edge  of  rule 
below  D.  It  should  be  noticed  that  the  accuracy  obtainable  by  using  this 
scale  is  not  as  great  as  in  the  method  given  above,  but  it  is  sufficient  for  many 
purposes,  being  approximately  ^  of  1%. 


—  18  — 

1   ^jQ    *~~*ft?Fl 

^^^^®^^t 

^^P^^f^^/^i^^P^Tf^'^t  *— j'ft^^g^''*'*^*"**'^**''*^^''^^^'^^^^^*1**1      *  *  i^****^***9  PB  ^»#^^^". 

l^^^kl^^^^W^t     KEUFFEL  &   ESSER  CO.,  NEW  YORK 

XL    CUBE  ROOTS. 

In  finding  the  cube  roots  of  numbers,  the  slide  may  be  used  either  direct 
or  inverted,  and  in  either  case  the  process  is  the  reverse  of  finding  the  cube  with 
the  corresponding  position  of  the  slide.  We  must  in  general  recognize  three 
cases.  Take  for  example  the  problem  of  finding  the  cube  roots  of  the  following: 
270,  27.0  and  2.70.  These  will  serve  to  outline  the  process.  To  find  y/270 
with  the  scale  direct:  R  to  270i?A,  slide  B  and  C  to  the  left,  using  thereby 
the  right-hand  scale  onJB  until  the  same  number  onB  appears  under  the  runner 
that  is  seen  on  D  under  the  right-hand  index  on  C.  When  6B  is  under  R,  we 
read  6.7  on  D,  and  when  6.46S  is  under  the  runner,  we  read  6.46  on  D  under  1C. 
Therefore,  6.46  is  v^TOT  To  find  ^27^0,  set  R  to  2.70LA,  and  using  the  same 
method  with  RB,  we  read  3  on  B  under  R  and  also  3  on  D  under  \C.  Using 
RA  and  LB,  we  arrive  at  the  same  result*.  To  find  \/2X,  set  R  to  2.7LA  and 
move  slide  to  the  right  so  as  to  use  LB.     1.39  then  appears  as  the  v/2.7 

Tabulating,  the  following  are  evident: 

tf270  =6.46     Using  right-hand  scales. 

\/2n^ =3.00        "        "     A  with  left  B  or  left  A  with  right  B. 

^2770=1.39         "     left-hand  scales. 

This  general  rule  can  then  be  formulated:  If  a  number  is  greater  than 
unity,  point  off  periods  of  three  places  to  the  left  of  the  decimal  point;  and  if 
there  are  no  digits  to  the  left  of  the  last  period,  use  the  two  right-hand  scales; 
if  one  digit  remains  to  the  left,  use  the  two  left-hand  scales ;  if  two  digits  remain 
to  the  left,  use  the  right  or  left  with  the  left  or  right.  With  a  number  greater 
than  unity,  in  most  cases  it  is  easier  roughly  to  approximate  the  result  and  use 
the  combinations  of  scales  giving  this  result.  To  find  the  cube  root  of  31700 
by  rule,  point  off  31 '700  and  we  have  two  digits  to  the  left,  and  the  rule  calls 
for  left  with  right  or  right  with  left.  Using  the  former  we  get  31.8.  The  cube 
roots  of  fractions  presents  greater  difficulty.     Making  a  table  as  above  we  have: 

^2.7     =  1.39     Using  the  two  left-hand  scales. 

#27     =     .646       "         "     "    right  " 

^\027    =    .30         "         "    right  with  the  left  or  vice  versa. 

#0027  =     .139       "         "        two  left  -hand  scales. 

This  leads  to  a  rule  for  fractions  as  follows:  Point  off  the  number  of 
zeros  immediately  to  the  right  of  the  decimal  point  in  groups  of  three,  counting 
the  decimal  point  as  one  zero;  if  there  are  no  zeros  immediately  to  the  right  of 
the  last  group,  use  the  two  left-hand  scales;  if  one  zero  remains,  use  the  two 
right-hand  scales;  if  two  zeros  remain,  use  right  with  left  or  vice  versa.  To 
find  the  cube  root  of  .000000738,  pointing  off  .00'000'0738,  we  have  one  zero 
remaining;  so,  by  the  rule,  we  use  two  rights  and  read  904,  and  as  we  have  two 
groups,  .00904  is  the  answer. 

It  should  be  distinctly  understood  that  these  rules  are  not  intended  to 
be  memorized,  but  a  study  of  them  will  simplify  cube  root  computations.  As 
already  suggested,  if  the  number  is  greater  than  unity,  a  method  of  approxima- 
tion and  trial  is  most  satisfactory. 


—  19  — 


With  fractions,  if  a  rule  be  desired,  the  following  is  simple:  Count  off 
the  zeros  between  the  decimal  point  and  the  first  significant  figure  of  the  number 
in  the  following  way  (counting  the  decimal  point  as  one  zero):  Right,  either, 
left,  right,  either,  left,  etc.,  the  scales  to  be  used  being  determined  by  the 
word  falling  on  the  last  zero.  As,  for  instance,  in  the  following:  .000386. 
Counting  as  indicated  above,  we  use  the  two  right-hand  scales,  and  read 
.0728  as  the  cube  root  desired. 

In  finding  the  cube  root  with  the  slide  inverted,  the  rules  given  above 
regarding  the  proper  reading  of  the  scales  hold.  Suppose  we  wish  to  find  the 
cube  root  of  1734;  evidently  the  result  lies  between  ten  and  twenty.  Set  R 
to  1734LA  and  place  the  middle  index  on  C  to  R.  Find  a  value  on  C  such  that 
the  number  on  the  same  point  on  D  is  identical.  Reading  on  RC,  we  find  im- 
mediately above  2585  on  D,  2585  on  C.  Obviously,  this  cannot  be  the  cube  root 
sought,  so  reading  now  LC  we  read  1202,  so  12.02  is  the  root  sought.  Again 
taking  the  same  example,  we  might  have  applied  the  rules  as  given.  Pointing 
off  1'734,  one  digit  remains,  and  the  two  left-hand  scales  are  called  for.  These 
are  the  ones  which  we  have  used  and  the  result  will  be  as  before.  Notice  if 
the  number  had  been  17340,  we  would  have  used,  as  we  did  in  the  first  trial,  the 
LA  with  the  RC  and  found  25.85,  which  would  have  been  correct  had  the  number 
been  17340. 

Take  for  another  example  the  cube  root  of  .000892 ;  counting  according 
to  the  rule  given,  we  end  on  the  word  "right"  and  we  find  .0963  as  the  correct 
answer.  In  finding  the  cube  roots  with  the  slide  inverted,  the  runner  should  be 
used  in  finding  the  coincidences,  as  it  helps  the  eye  materially  and  is,  of  course, 
not  needed  at  the  center  index  of  C  after  the  placing  of  that  index. 

We  can  now  solve  problems  of  the  type  f  (84)2.  With  the  slide  direct, 
set  R  to  84D  and  move  the  slide  until  the  same  number  onB  is  under  R,  that  is 
seen  on  D  under  the  left  index  of  C;  in  this  case,  using  the  left  scale  onB,  we 
read  413.  Notice  that  802  =  6400  and  that  41.3  cannot,  therefore,  be  the  number 
sought;  however,  if  we  use  RB  we  find  192  and  19.2  is  the  correct  answer.  The 
question  of  which  slide  to  use,  direct  or  inverted,  is  largely  a  personal  one.  The 
operator  should,  however,  master  both  methods. 


THE  POLYPHASE  SLIDE  RULE. 

With  the  Polyphase  rule  cube  roots  may  be  found  to  an  accuracy  of 
14  of  1  %  by  the  use  of  the  scale  on  the  edge  of  the  rule.  It  will  be  noticed  that 
the  scale  is  divided  into  three  parts.  The  left-hand  scale  being  for  numbers  of 
one  digit  to  the  left  of  decimal  point,  the  middle  scale  for  numbers  of  two 
digits  to  the  left  and  the  right-hand  scale  for  numbers  of  three  digits  to  the 
left  of  the  decimal  point.  If  the  number  of  digits  be  greater  than  three,  sub- 
tract any  multiple  of  three  and  use  the  scales  as  indicated  above;  thus,  for  the 
cube  root  of  1,860,000,  subtract  6  from  7  and  find  cube  root  of  1.86  on  scale 
to  left,  setting  runner  to  number  on  the  cube  scale  and  reading  the  answer 
on  D  as  1.23. 

It  will  be  seen  that  the  cube  scale  makes  problems  of  the  type  At  very 
simple.  Set  R  to  A  on  the  cube  scale  and  read  the  answer  on  A  under  R. 
10s  =  4.65.     Set  R  to  10  on  cube  scale  and  read  4.65  on  A  under  R. 


—  20  — 


XII.     LOGARITHMS. 

Logarithms  of  numbers  may  be  found  on  the  slide  rule  in  two  ways. 
First,  set  LLC  to  the  number  on  D  and  obtain  the  mantissa  of  the  logarithm 
on  the  center  scale  of  equal  parts  to  be  found  on  the  back  of  the  slide,  by 
reversing  the  rule  and  reading  under  the  line  etched  on  the  xylonite  on  the  left 
end  of  the  back  of  the  rule.  To  find  the  logarithm  of  265,  set  1C  to  265D  and 
reading  on  the  center  scale  on  the  back  of  the  rule  under  the  line,  we  find  423. 
The  desired  logarithm  is  then  2.423.  It  should  be  remembered  that  the  number 
found  on  the  rule  gives  only  the  mantissa  of  the  logarithm  and  that  the 
characteristic  must  in  every  case  be  supplied  in  order  to  complete  the  logarithm. 
In  the  second  method,  we  reverse  and  invert  the  slide  and  read  the  mantissa 
on  the  center  scale  directly  over  the  number  on  D;  the  runner  being,  of  course, 
used  to  do  this.     Thus  we  find  the  logarithm  of  265  to  be  2.423  as  before. 

In  finding  the  numbers  corresponding  to  logarithms,  the  reverse  of 
either  of  these  two  methods  is  employed;  the  mantissa  alone  being  set  upon 
the  scale.  Example:  to  find  the  number  corresponding  to  the  logarithm 
3.697:  With  slide  direct,  set  697  on  center  scale  on  back  of  slide  to  line  on 
xylonite  on  back  of  rule,  and  read  answer  on  D  under  1C,  4980;  or  reverse  and 
invert  the  slide,  put  the  indices  in  coincidence  and  set  runner  to  697  on  the 
scale  in  center,  and  read  under  runner  4980,  the  answer,  on  D. 


XI11.     GENERAL  INVOLUTION  AND  EVOLUTION. 


Logarithms  are  used  in  slide  rule  work  mainly  to  raise  numbers  to 
fractional  powers.  In  raising  numbers  to  integral  powers,  we  can  use  either  the 
logarithmic  method  or  a  combination  of  squaring  and  cubing  with  direct 
multiplication.  Thus:  (317)7  =  [(3.17)2]2  (3.17)3 or  (3.17)2X  (3.17)2X  (3.17)3. 
This  method  is  applicable  to  any  power  or  root.  Suppose  we  solve  the  problem 
first  by  the  method  outlined  above,  with  this  change  (3.17  X  3.17)2  X 
(3.17)3  =  (3.17)7.  Slide  direct;  1C  to  3.17D,  R  to  3.17B,  LB  to  R,  R  to  10LB. 
Read  answer  under  Ron  A,  3210.  The  logarithmic  method  depends,  of  course, 
upon  the  fact  that  if  we  multiply  the  logarithm  of  the  number  by  the  exponent, 
.the  answer  is  the  logarithm  of  the  required  number.  Set  \C  to  3 . 1  ID.  Read  on 
center  scale  on  back  of  slide  .501,  the  characteristic  being  0.  Set  1C  to  5011), 
R  to  1C.  Read  on  D  under  R  3.505.  Set  505  on  center  scale  on  back  to  mark 
on  xylonite  and  read  under  1C,  3210  on  D,  which  is  the  answer.  It  may  be 
easily  seen  that  the  first  method  is  the  more  accurate,  while  for  the  seventh 
power,  at  least,  the  second  is  much  the  shorter.  The  logarithmic  method  is 
especially  adapted  to  problems  involving  fractional  powers  or  roots,  except 
where  the  desired  root  or  power  is  a  multiple  of  two  or  three.  For  instance, 
suppose  we  wish  to  find  (717)1,  or  as  it  might  be  stated,  the  fifth  root  of  717 
cubed.  We  may  cube  717  and  extract  the  fifth  root.  Cubing  717  gives 
370000000,  and  the  logarithm  of  this  number  is  by  the  rule  8.568.  Dividing 
by  5  gives  1.713  and  the  number  corresponding  to  the  mantissa  .713  is  516,  so 
the  correct  answer  will  be  51.6.  An  easier  solution  is  this:  Divide  3  by  5 
giving  .6;  the  logarithm  of  717  is  2.855;  2.855  X  .6  =  1.713  and  the  number 


—  21  — 


corresponding  as  before  to  the  mantissa  .713  is  51.6.  The  problem  as  outlined 
above  is  actually  worked  out  as  follows :     Mentally  —  =  .6;  set  \C  to  717D, 

o 

read  under  line  on  back  of  rule  855,  and  with  characteristic  2  the  logarithm  is 
2.855;  set  IRC  to  2.855D  and  read  on  D  under  6C  the  number  1.713;  set  713 
on  the  center  scale  on  the  back  of  the  slide  to  the  line,  and  read  on  D  under 
1C,  516,  and  as  the  characteristic  was  1  the  answer  will  be  51.6.  Remember 
that  the  rule  in  every  case  gives  only  the  mantissas  and  that  the  characteristic 
must  be   supplied   independently. 

The  problem  of  finding  the  root  of  numbers,  the  index  of  the  root  being 
fractional,  is  slightly  more  complicated.  Solve  •'v/L~9  as  an  example:  This 
is  equivalent  to  (1.9).oV  The  value  of  1/.03,  determined  by  the  method  of 
finding  reciprocals  as  previously  explained,  is  33.3.  The  logarithm  of  1.9  =0.279; 
0.279  X  33.3  =  9.3;  number  corresponding  to  mantissa  .3  is  2.  So  the  answer 
will  be  2  000  000  000. 

As  another  type,  take  (.00273) -232,  which  represents  the  most  difficult 
class  of  problems  in  involution.     Find  logarithm  of  .00273  to  be  7.436 — 10  or 
— 3.436,  remembering  that  the  characteristic  is  negative  and  the  mantissa 
positive.     -3.0  X  .232  =  -.696 
.436  X  .232  =.101 

-.595  adding  and  subtracting  10,  the  logarithm  is 
9.405-10;  and  the  number  corresponding  to  mantissa  .405  is  254.  Therefore, 
(.00273)-232.  =  .254 

Let  us  now  check  this  result  as  follows: 
(.00273)*232    =  .254  if  the  work  is  correct. 

(.00273)7x3=  .254,  for  we  set  232C  to  ID  and  under  3C  find  13D. 
(.00273)3  should  equal,  if  the  work  performed  is  correct,  (.254)13. 
(.00273)3  =  .0000000203. 

=  -ir(.254)2]^  3  X  .254. 
=  .0646. 
=  .00416. 


(.254) 13 
(.254)2 
(.0646)2 
(.00416)3  =  .000000072. 
.000000072  X  .254  =  .0000000183c/ifc. 


THE  POLYPHASE  SLIDE  RULE 

The  methods  of  squaring  and  cubing  having  been  explained,  it  is  easy 
to  see  how  to  raise  a  number  directly  to  the  4th,  5th  and  6th  power  by  the  use 
of  CI  scale. 

a4,  aCl  to  aD.    Answer:    A  over  LB. 

a5,  aCl  to  aD.    Answer:    A  over  aB. 

a6,  aCl  to  aD.     Answer,  on  cube  scale  under  1C 

XIV.  EQUATIONS  INVOLVING  FRACTIONAL  EXPONENTS. 

Equations  of  the  type  X«  =6  may  be  solved  by  the  methods  already 

given,  for  XN  =  b  is  equivalent  to  X  =  6n  .  Now  let  us  consider  a  few  examples 

occurring  in  engineering  work.    The  volume  of  the  compression  space  in  the 

cylinder  of  a  gas  engine  is  \i  of  the  piston  displacement.    What  is  the  efficiency 

(Vd\k-\ 
=  \Ve) 


of  the  engine?  The  formula  to  be  used  is  E 


where  Vd  =  \  Vc  and 


—  22 


k  =  1.405.  Substituting  these  values  in  the  formula,  E  =  (.25).405.  Set  \C 
to  25D  and  read  under  line  on  back  of  the  rule  on  the  center  scale  398,  so  the 
logarithm  required  is  -1.398,     -1  X  .405  =  -.405 

.398  X  .405  =     .1609,   IRC  to   .405D,  read 

under  398C.     Next  subtract  

-.244 
-.244  from  10-10,  and  we  have,  as  the  logarithm  of  the  answer,  9.756-10. 
Set  756  on  the  logarithmic  scale  on  the  back  of  the  rule  and  read  on  D  under 
1C,  574;  E  =  1-.574  or  about  43% 

Take  as  another  example,  Hodgkinson's  formula  for  the  strength  of 
cast-iron  columns  the  lengths  of  which  exceed  thirty  times  their  diameters. 

J)3-55 

The  formula  is  W  =  99318 ;  where  W  equals  the  breaking  load;  D  the 

£1-7 

diameter  of  the  column  in  inches;  andB  its  length  in  feet.     Let  D  =  3";  and 

3.3-55 

B  =  11.5';    W  = X  99318.     The  example  may  be  solved  by  finding 

11.5»'» 
the  numerator  and  the  denominator  of  the  fractions  separately    and   then 
dividing  them. 

Log.  3  =  .477  Log.  3355  =  .477  X  3.55  =  1.695  Number  corres.  =  49.5 
Log.  11.5  =  1.061      Log.  11.51"   =1.061  X  1.7  =  1.805       "        "        =  63.8 

49.5 
So  W  =  99318  X  =  76900. 

63.8 

This  method  is  quite  simple,  but  the  following  in  many  cases  will  prove 

to  be  more  satisfactory: 

33.55 

We  may  write  the  formula:     W  =  99318 .        Since  3*  =  11.5, 

(32-22)1.7 

Log.  11.5  1.061  33'58X99318 

and   x    = = =  2.22;  then  we  have =   W   => 

Log.  3  .477  3377 

33-55  1 

99318  X =  99318  X ;  and  since  Log.  3*22  =  .477  X  .22  =  .105  and 

33.77  322 

99318 

the  number  corresponding  =  1.27,  we  have  W  = =  77000.     This  will 

1.27 

cover  most  of  the  cases  which  arise  in  involution  and  evolution. 

XV.    TRIGONOMETRIC  COMPUTATIONS. 

Trigonometric  functions  may  be  found  on  the  rule  either  with  slide 
direct  or  reversed;  sines  and  tangents  being  given  directly  by  the  readings  on 
the  rule.  On  the  reverse  of  the  slide  will  be  found  two  scales,  one  at  the  top 
marked  S  giving  sines,  and  the  lower  marked  T  giving  tangents.  To  find 
the  sine  of  an  angle  with  the  slide  direct,  set  the  angle  on  scale  S  to  the  line  on 
xylonite  on  the  back  of  the  rule,  and  read  the  sine  of  the  angle  on  B  under  the 
index  on  scale  A.  For  example,  to  find  the  sine  of  15°,  we  set  15  on  scale  S, 
to  the  line  and  under  middle  or  right  index  of  A  we  read  .259;  if  we  wish  the 
tangent  of  15°,  we  set  15°  on  scale  marked  T'to  the  line  on  the  back  of  the  rule 


—  23 


and  read  on  C  under  index  on  D,  .268.  Or  the  slide  may  be  reversed  —  the 
left-hand  indices  brought  into  coincidence,  and  the  sines  and  tangents  read 
directly,  using  scale  of  sines  with  A  and  scale  of  tangents  with  D.  Reading 
on  A  over  15°  onB  we  have  .259,  and  reading  on  D  under  15°  on  C  we  find  .268. 
Notice  that  if  the  sine  occurs  on  the  scale  LA,  one  zero  stands  between  the 
decimal  point  and  the  first  significant  figure  of  the  desired  sine;  the  sine  of  2° 
is  .0349  and  the  sine  of  20°  26'  is  .349.  This  should  be  kept  clearly  in  mind.  In 
the  same  way,  all  the  values  of  the  tangent  to  be  read  on  scale  D  will  lie  between 
1.  and  .1.  The  sines  and  tangents  of  small  angles  are  practically  identical,  and 
may  be  found  in  one  of  two  ways :-  either  by  using  the  gauge  point  on  the 
rule,  or  by  an  application  of  the  proposition  that  the  sines  and  tangents  of  very 
small  angles  are  proportional  to  the  angles  themselves.  Using  the  first  method, 
let  us  find  the  sine  of  4'  and  56".  Set  the  gauge  point  which  is  seen  just  before 
the  2°  mark  on  the  scale i?  to  4A  and  read  on  A  over  LB,  1162.  Then  set  gauge 
point  which  is  near  1°  10'  to  56A  and  read  on  A  over  LB,  2715.  Remembering 
that  the  sine  of  1"  is  .000005  and  that  the  sine  of  1'  is  .0003,  we  will  write  our 
first  reading  .001162  and  add  to  that  our  other  reading  0.002715  and  obtain 
,0014335,  which  is  either  the  sine  or  tangent  of  4'  56". 

The  second  method  is  as  follows:  The  sine  of  4'  56"  is  equal  to  1/10  the 
sine  of  40'  560"  =  1/10  sine  49.33';  and  by  the  rule  we  find  sine  49'  33"  = 
.0143  and  1/10  of  this  value  is  .00143.  Suppose  we  multiply  4'  56"  by  20,  we 
have  sine  4'  56"  =  1/20  sine  1°  38.66'.  We  find  sine  4'  56"  =  1/20  =  .0286 
=  .00143;  the  latter  method  is  preferable  in  almost  all  cases.  As  will  be 
noticed,  we  cannot  find  directly  the  sine  of  angles  between  90°  and  360°,  nor 
tangents  of  angles  greater  than  45°.  All  that  is  necessary  is,  of  course,  to  reduce 
the  functions  as  expressed  to  functions  of  an  angle  less  than  90°.  Thus  the  sine 
of  169°  =  sine  11°  by  the  familiar  rule  of  trigonometry,  and  the  tangent  of 

75°  = i-r-n*     In  cases  where  we  have  to  deal  with  other  functions  than  the 

tan  15 

sine  or  tangent  we  have  to  transform  the  function  by  trigonometry  until  the 
required  function  is  expressed  in  terms  of  the  sine  or  the  tangent.  Cosine  A  = 
sine  (90°-A);  that  is,  to  find  the  cosine  of  49°  13'  we  find  the  sine  of  40°  47', 
which  by  the  rule  is  .653.     This  method  enables  us  to  find  all  the  functions 

very  simply:    cos  A  =  sin  (90°-A),  cot  A  =  r t->  sec  A  = T  esc  A  = 

J        * J  v  tan  A  sin   A 

-,or  in  finding  the  cosine  of  A  we  might  divide  the  sine  of  A  by  the  tangent 


sin  A 

of  A.    Thus  to  find  the  cos  30°  15',  find  .503  as  the  sine  of  30°  15'  on  scale 

A,  set  R  to  503D,  30°  15'  C  to  R  and  read  under  1C  .864  on  D,  which  is  the 

cosine  required.     In  general,  the  first  method  is  easier  when  the  ang  e  lies 

between  45°  and  90°,  and  the  second,  when  the  angle  lies  between  0°  and  45°. 

Let  us  consider  the  example  of  finding  the  cotangent  of  75°;  evidently  cot 

75°  =  tan  15°.     Notice  that  with  the  slide  reversed  we  can  multiply  directly 

by  the  sine  or  tangent  of  the  angle  without  first  determining  the  actual  numerical 

value  of  the  function;  needless  to  say,  the  process  of  division  is  equally  simple. 

217  3 
Solve  —  ,  '•'  ,  .,',  set  13°  14'C  to  217.3D  and  reading  under  1C  find  on  scale 
tan  13    14 

D  923.0  as  the  answer.     Solve  653  X  sin  36°  41';  set  LB  to  653A,  R  to  36°  41'.B 

and  read  answer  390  on  A  under  R. 


—  24  — 


1.36 

Required  the  solution  of  the  right  triangle  as  given  above.  With  scale  direct 
divide  2.17  by  4.36  and  find  .497  as  the  result.  Let  the  runner  rest  at  .497  and 
reverse  the  slide,  set  indices  together  and  read  on  C  under  R,  26°  25',  which  is 
the  value  of  the  angle  opposite  the  shorter  leg;  the  other  angle  will  be  63°  35'. 
Set  R  to  2.17 A,  26°  25'B  to  R  and  read  on  A  over  IB  4.86,  which  is  the  length 
of  the  hypothenuse.  Check  this  result  by  squaring  2.17  =  4.71,  and  squaring 
4.36  =  19,  adding  the  two  =  23.71,  extracting  the  square  root  =  4.86  chk. 

We  may  solve  an  oblique  triangle  using  the  law  of  sines  very  simply,  in 
any  case  where  that  law  applies. 


Solve  the  following  oblique  triangle: 


4.32 


34° 40' 


With  the  slide  reversed,  set  34°  40' B  to  3.19A  and  read  the  answer  onB  under 
4.32A,  as  50°  35',  which  is  the  value  of  the  angle  opposite  the  side  4.32.  The 
other  angle  will,  of  course,  be  94°  45'  and  since  we  know  sine  94°  45'  =  sine 
85°  15',  we  can  easily  find  the  side  opposite,  by  a  simple  proportion  as  above. 

In  general,  it  can  be  stated  that  any  formula  adapted  to  logarithmic 
computations  is  adapted  to  computations  with  the  slide  rule.  If  a  particular 
problem  calls  for  logarithmic  functions,  they  can  be  very  simply  found  by  finding 
the  logarithm  of  the  function.  Suppose  log  tan  38°  is  desired.  Reverse  the 
slide  and  set  the  runner  on  38°  C,  then  invert  the  slide  reversed  as  it  is,  and 
read  on  the  scale  of  logarithms  under  the  runner  .893,  which  is  the  required 
mantissa,  the  complete  logarithm  being  9.893-10.  In  finding  the  log  sine  use 
the  same  general  process,  but  multiply  the  mantissa  as  obtained  on  the  scale 
by  2,  and  take  the  fractional  remainder  as  the  required  mantissa.  Find  the 
log  sin  20°.  With  the  slide  reversed  and  the  indices  coinciding,  set  R  to  20° B, 
invert  the  slide  and  read  on  the  logarithmic  scale  under  R,  767.  Multiplying 
by  2  gives  1.534  and  .534  is  the  mantissa  required.  The  logarithmic  functions 
are  often  useful  in  checking  logarithmic  work. 


—  25  — 


XVI.    VARIATION  AS  THE  SQUARE. 

The  following  example  is  typical  of  a  wide  variety  of  problems  to  which 
the  slide  rule  offers  a  simple  solution.  The  example  as  given  is  very  simple,  but 
from  it  others  may  be  deduced.     Given  the  parabola  as  shown: 


OD  °    5ft. 
OL  '  18ft. 


to  find  the  distance  OA  which  corresponds  to  a  distance  of  AB  =  3.37  feet. 
Subtracting  3.37  from  5  gives  1.63  =  CK.  To  solve,  set  18C  to  5A,  R  to  1.63A, 
and  read  answer  under  R  on  C  as  10.29.    This  may  be  checked  by  finding  the 

324 

equation  of  the  parabola,  which  is  found  to  be  y2  = x;  when  x  =  CK  and 

5 


y  =  OA,  we  have  y  = 


V 


324  X  1.63 


=  10.29.  Problems  of  this  type  are  very 


useful  in  structural  design,  where  the  moment  curve  is  a  parabola,  or  in  rail- 
roads where  a  vertical  curve  is  parabolic  in  form.  The  setting,  as  given,  is 
deduced  from  the  fact  that  on  a  parabola  the  abcissae  are  proportional  to  the 
squares  of  the  ordinates.     Let  us  work  one  simple  problem  in  detail. 

Given  a  beam  uniformly  loaded  with  a  load  of  1000  lbs.  per  foot ;  the 
beam  being  17  feet  long,  to  find  the  moments  at  one  foot  intervals  over  the  span: 
in  this  case  the  moment  curve  is  a  parabola  with  vertical  axis.     Center  distance 
WL*      1000  X172  17 

= = ;  SB  to  17D,  read  the  answer  on  A  over  LB  as  36100;  set  —  or 

8  8  2 

8.5C  to  3.61A  and  corresponding  to  the  following  distances  on  the  beam,  as 
measured  from  the  center,  we  have  as  per  the  table: 

8.5         7.5         6.5         5.5        4.5        3.5      2.5      1.5      0.5 
36100     28200     21200     15100     10100     6130     3130     1120     125 

These  values  are  obtained  by  setting  as  above  and  reading  the  numbers 
on  A  corresponding  to  distances  on  C.  That  is,  8.5C  corresponds  to  36100A, 
7.5  to  28200A,  etc.  To  find  the  moment  subtract  each  number  as  found  from 
36100  and  we  have  the  actual  moments  at  corresponding  points. 

End     123  45  678 

0       7900        14900       21000       26000         29970       32970       34980       35980 

The  distances  being  expressed  now  as  distances  from  the  end  of  the  beam 
instead  of  from  the  center. 


—  26 


XVII.     SOLUTIONS  OF  SPECIAL  FORMS. 
The  following  solutions  of  general  formulae  are  instructive. 

(1).     The  type  — ;  which  occurs  in  hydraulic  computations. 
gx 

Let  v  =  31,  g  =  32,  x  =  8. 

Mental  approximation;  31  divided  by  8  is  approx.  =  4. 

Set  32B  to  31D,  R  to  IB,  8B  to  R,  read  answer  on  A  over  IB,  3.75. 


(2).     The  type 


J-1 


is  solved  in  the  same  way  except  that  the  answer 


is  read  on  D  instead  of  A.     Care  should  be  taken,  however,  in  all  problems 
involving  roots  to  use  the  proper  scales,  when  using  A  andB  together. 

m  v2 

(3).     The  type ;  same  as  (1),  with  a  coefficient. 

2g 
Let  v  =  8.5,  m  =  .93,  2g  =  64.4 
Mental  approximation:     80/64  =  approx.  1. 

Set  64.4J5  to  8.5D,  and  over  .935  find  1.04  answer  on  A. 

(4).     The  type  ^2gh  ; 
a 

Let  2g  =  64.4,  h  =  12.5,  a  =  5. 

Mental  approximation;  — - —  =  approx.  5.0. 

5 

Invert  the  slide;  set  64.4C  to  12. 5A,  under  5B  find  answer  on  D  as  5.67. 

Care  should  be  taken  to  use  the  proper  scales;  notice  that  with  the  wrong 

choice  179  is  read  under  5B. 

(5).    The  type        b    ; 
c 

Find  the  square  root  of  b,  divide  by  c  and  multiply  by  a.     This  can  be 

done  either  direct  or  inverted.     Direct:  set  cC  to  bA,  and  under  aC  find  the 

answer  on  D. 


(6).    The  type  a 


bXc 


;take  for  an  example  of  this  type  the  formula 


\      d 
for  the  flow  through  a  nozzle  as  expressed  by  the  formula:     g  =  .7854d2 


V 


2gh 


^-^prv  and  let  us  consider  its  solution  in  detail.     First  find  the  value 

of  .7854d2,  which  is  the  area  of  a  circle  with  diameter  d. 
Let  d    =1|  in.,  then  .7854d2  =  2.4  sq.  in. 
Let  D  =  Sy2  in.,  2g  =  64.4,  c  =  .97,  h  =  130. 

(^)2  =  1.06;      (^)2  =  .25    [(1)2-(|)2]  =  .81.     The  formula  has 


now  been  reduced  to  2.4 


-JL- and  this  formula  is  of  the  type^     Xc  .  Mental 

^1  .81  "V   d 


approximation:  65X150  =  10000,1/10000  =  100,     100  X  2.4  =  240. 


the  process 


Considering  this  formula,  it  is  evident  that  when  we  have  completed 

4 


IbXc 


the  result  should  be  indicated  upon  scale  A  in  order  to 


facilitate  the  finding  of  the  square  root;  this  being  found,  we  multiply  the  result 
by  a.  Any  method  of  multiplication  which  gives  the  result  on  A  can  be  used. 
Theie  is  no  reason,  therefore,  why  the  scale  should  not  be  used  direct. 

Set  SIB  to  64.4A,  R  to  130.B,  IB  to  R,  under  2.4C  read  the  answer  on  D  as 
244  cubic  feet  per  second.  Again  we  must  be  careful  in  selecting  the  scales  to 
use  on  A  and  B.  Estimating  the  value  of  the  expression  under  the  radical  and 
mentally  extracting  the  square  root,  it  is  seen  to  be  about  100,  therefore,  the  re- 
sult of  the  quantity  under  the  radical  should  appear  on  the  left-hand  scale  of  A. 
This  simple  approximation  should  always  be  made  where  a  root  is  involved. 

m       \/H* 

(7).    The  type  —  or .     The    analysis    shows    that    we    should 

b  b 

cube  H  so  that  the  result  falls  on  A,  transfer  to  D  and  divide  by  b. 

Let  H  =  1.79,  b  =  3.32.  Notice  that  H3  =  approx.  8,  so  we  wish 
the  result  of  this  step  to  fall  on  LA. 

Mental  approximation:     3/3.3  =  .9. 

Slide  direct:  set  1C  to  179D,  R  to  1795,  332C  to  R,  read  answer  on  D 
under  1C,  as  .722.  With  slide  inverted,  set  179C  to  179D,  and  under  332E  find 
answer  on  D  as  .722.  As  the  second  method  requires  a  change  of  index,  in 
this  particular  case  it  is  little,  if  any,  shorter.  In  most  cases,  however,  the  second 
method  is  the  quicker  of  the  two,  and  is,  therefore,  to  be  recommended. 

(8).  The  type  a  X  bi ;  an  example  of  this  type  is  the  formula Q  =  3.336/tl 
due  to  Francis.  The  quantity  Q  is  the  discharge  over  a  trapezoidal  weir  in 
cubic  feet  per  second,  b  the  length  on  the  crest  in  feet,  and  h  the  head  of  the 
weir.  The  formula  may  be  written  &/.3  X  fcf;  as  is  readily  seen,  the  inverted 
slide  offers  the  best  solution. 

Let  h  =  1.43,  and  b  =  4.51. 

Mentally  approximate:     (l^)3  =  3.5,  V3.5  =  2,  2  X  5  X  3  =30.  Ans- 

Solution  with  slide  inverted:      Set  1.43C  to  1.43D,  R  to  1C,  4.5LB  to  R, 
read  answer  on  D  under  3J5  as  25.7.    Notice  that  the  cube  of  1.43  is  approxi- 
mately 3.5,  so  that  the  result  of  the  process  of  cubing  should  appear  on  the 
left  scale  of  A.     If  it  is  made  to  appear  on  the  right,  the  answer  is  found  to 
be  813,  which  is  impossible. 

VaX  b 
(9).    The  type is  sometimes  given.    This  is,  of  course,  equiva- 


lent to  the  expression 


x 


«X62 


and  in  this  form,  it  is  of  more  common  occurrence. 


c- 


Its  solution  requires  us  to  find  the  ratio  — ,  square  it  and  multiply  by  a. 

c 

Let  a  =  9.18,  b  =  3.12,  c  =  4.26. 

Mental  approximation:     81/16  =  5. 

Use  the  slide  direct;  set  4.26C  to  3.12D,  read  answer  on  A  over  9185 
as  4.93.  These  types  shown  are  few  in  number,  but  they  serve  to  illustrate  the 
method  of  finding  a  setting  for  a  given  formula. 


-  28 


XVIII.     GENERAL  CONDITIONS. 

Remember  in  general  that  all  formulae  should  be  expressed  in  the  form 
of  products,  as  the  first  step  in  their  solution  by  means  of  the  slide  rule.  The 
expression  for  the  cosine  of  any  angle  of  a  triangle  in  terms  of  its  sides  is  a 


familiar  example  of  this  process.     Cos  3^A 


^ 


s   (s-a) 


be 


where  s 


the 


sum  of  the  sides  a,  b,  c. 
simple  transformation. 


This  expression  is  derived  from  the  law  of  cosines  by  a 
As  is  readily  seen,  the  form  above  is  adapted  to  slide 


rule  computation,  while  the  law  of  cosines  is  not.     That  is,  Cos  A  = 


b2+c2 


26c 


cannot  be  readily  solved  by  either  the  slide  rule  or  by  logarithms. 

It  is  sometimes  helpful  to  add  and  subtract  numbers  on  the  rule  when 
paper  and  pencil  are  not  at  hand.  This  may  be  done  by  reversing  and  inverting 
the  slide;  suppose  we  have  the  following:     Add    3.17 

5.36 
-3.81 
-7.42 
6.89 
—4.19 
With  the  slide  reversed  and  inverted,  bring  indices  into  coincidence, 
R  to  3.17  log  scale,  0  log  scale  to  R,  R  to  5.36  log  scale,  3.81  log  scale  to  R,  R  to 
0  log  scale,  7.42  log  scale  to  R,  R  to  0  log  scale,  invert  slide  and  set  right-hand 
indices  into  coincidence.     Read  the  distance  on  log  scale  between  runner  and 
6.89.     Read  in  this  way:     3     1.1     .09  or  4.19.     The  operator  will  easily  grasp 
the  principle  of  this  setting  and  can  readily  see  in  what  cases  its  application 
will  be  of  advantage. 


—  29  — 


PROBLEMS. 

EQUIVALENT  RATIOS. 

'(1).     Find  the  areas  of  circles  the  diameters  of  which  are  respectively! 
7.93.  9.76,  .01345,  64.8,  and  33.71  in. 

Ansicers:     49.4,  74.8,  .0001421,  3300,  893. 

f  (2).     Find  the  number  of  cubic  feet  corresponding  to  the  following 
numbers  of  U.  S.  gallons:     8.0,  9.78,  351,  1003. 

Answers:     1.070,  1.307,  46.9,  134.1. 

(3).     Transform   the   following   volumes   expressed   in    cubic   feet   to 
litres:     33.9,  67.8,  85.4. 

Answers:     960,  1921,  2420. 

*  (4).     Find  decimals  of  inches  corresponding  to  the  following  fractions 
of  inches:     1/8,  3/32,  5/16,  9/16,  17/64,  19/64. 

Answers:     .125,  .0938,  .313,  .563,  .266,  .297. 

(5).     Find  decimals  of  a  foot  corresponding  to  the  following  numbers 
of  inches:     3X,  5H,  7%,  9*A,  HM,  &H,  *H. 

Answers:     .260,  .438,  .646,  .792,  .979,  .698,  .375. 

.  (6).     Find  the  pressures  in  tons  per  square  foot  due  to  the  following 
depths  of  water:     6J4  9,  17,  31.5,  and  76  feet. 

Hint:     Determine  first  the  ratio  between  pounds  per  square  inch  and 
tons  per  square  foot. 

Answers:     .203,  .281,  .531,  .984,  2.38. 

(7).     Express  to  the  nearest  even  sixteenth  of  an  inch  the  following 
decimals  of  a  foot:     .938,  .642,  .733,  .657. 

Answers:     11  \i,  7,,/i6,  8"/i6,  7|. 

(8).     Transform  the  following  pressures  given  in  inches  of  mercury 
to  equivalent  pressures  expressed   as   atmospheres:     35.7,   33.4,   31.6,   27.3. 

Answers:     1.194,  1.117,  1.057,  .910. 

(9).     Find  the  weight  of  copper  strips  1  in.  wide,  1  ft.  long,  and  of  the 
following  thicknesses:     3/i6,  \i,  5A,^-%  in. 

Answers:     .716,  .953,  2.38,  4.28. 

(10).     Find  the  weight  in  grams  of  a  steel  plate  circular  in  shape  with 
diameter  .1125  in.  and  thickness  u/32  of  an  inch. 

Answer:     .759. 


30  — 


SQUARES  AND  SQUARE  ROOTS. 

(1).  Find  the  squares  of  the  following:  10.17,  .979,  126.5,  .0326, 
4.77,  87.9. 

Answers:     103.4,  .958,  16000,  .001063,  22.75,  7730. 
|  (2).     Find  the  square  roots  of  the  following:     .0735,  .520, 1391,  600,  3.17. 
Answers:     .271,  .721,  37.3,  24.5,  1.78. 

(3).     Find  the  squares  of  the  following:     43/52,  47/64,  89/71. 
Hint:     52C  to  43D,  read  answer  on  A  over  IB. 

24       8.9      '  27        34.5  110       100.5 

Answers:      .683  or  —  or  —  ;  .540  or  —  or ;  1.57  or or . 

35   13       50    64        70     64 

(4).  Find  the  square  roots  of  the  following  expressed  as  even  sixty- 
fourths:     9/64,  19/32,  3/8,  7/16,  1/32. 

Answers:    24/64,  49/64,  39/64,  42/64,  11/64. 

(5).  Find  the  length  of  the  side  of  a  square  to  the  nearest  64th  of  an 
inch,  the  square  containing  9.81  square  inches. 

200 
Answer:    — or  2>Y%. 
64 
■  (6).     Find  the  diameter  of  a  circle  to  the  nearest  thirty-second  of  an  inch, 
the  circle  containing  7.97  square  inches. 

Hint:     Use  the  ratio  found  on  the  reverse  of  the  rule  79: 70. 

204 
Answer:    —  or  33/i6. 
64 


MULTIPLICATION  AND  DIVISION. 

(1).     Solve  16.16  X  3.1416  X  18.13.     Answer:     921. 

39.8 
Solve .     Answer:     .852. 


(2). 

(3). 

(4). 

(5). 
(6). 


46.7 

83.7  X  1163 
Solve .     Answer:     12.54. 

77.6 

100.3  X  69.3  X  33.7 
Solve .     Answer:    2.75. 

88.2  X  54.6  X  17.70 
1131  X  94.8 
Solve .  Answer:     .1750. 

79.9  X  104.4  X  73.5 
Solve  24 


47 
18 
51 
1 
19 


Answer:    0.0761. 


—  31  — 


\J  (7).     Solve  the  following  proportion,  stating  the  answer  to  the  nearest 

17i  9|  54         22         11 

thirty-second:     =  — .     Answer:     x  =  —  or  1 —  or  1— . 

3£  x  32         32         16 

(3.18)2  x 

(8).     Solve  the  following: =  .     Answer:    x  =  3.27. 

(10.17)2        33.4 

(9).     Give  the  general  setting  for  the  solution  of  problems  of  the  type 

a2        x 
of  (8)  —  =  — .    Ansiver:    Set  bC  to  aD,  answer  on  A  over  cB. 
£2         c 

(10).    Find  the  answer  to  this  problem  in  kilogram  meters  when  the 
quantities  are  expressed  in  feet  and  pounds: 

3.1416  X  (5.07)2  X  189  X  2.73  X  550 

_ m  Answer:     791000. 

4 

Hint:  Deduce  from  the  table  of  equivalents  a  factor  to  satisfy  this  case 
and  multiply  the  expression  by  it. 

INVERTED  SLIDE. 

(1).    Find  the  reciprocals  of:    389,  40.9,  7.79,  1019,  6.59. 

Answers:    .00257,  .0244,  .1284,  .000981,  .1517. 

(2).     Multiply  13.14  X  9.63  X  7.46  X  349.     Answer:     329000. 

Multiply  159.6  X  .834  X  76.7  X  9.13.     Answer:     93200. 
(3).     Solve  this  example  with  the  slide  inverted  and  also  with  the 

3.43  X  79.6  X  928  X  .1119  X  .06888 

slide  direct:     .     Answer:     .0001471. 

19.15  X  47.7  X  344  X  42.2 

(4).     Find  a  simple  method  for  solving  the  following  and  determine 

1111 

their  numerical  values:    , , 

(3.29)2     (18.72)2   (4.13)2      (1.251)2 

Answer:  Invert  slide  and  read  over  3.29,  18.72,  etc.,  onJ5,  the  answer 
on  A  as:    .0924,  .00285,  .0586,  .638. 

(5).  Find  a  simple  setting  for  problems  of  this  class  and  determine 
numerical  results  with  the  data  as  given. 

A  lever  of  the  first  class  has  the  fulcrum  6  in.  from  the  end  where  rests 
a  weight  of  380  lbs.  At  what  distances  from  the  fulcrum  will  the  weights  given 
be  placed,  one  weight  acting  at  a  time,  in  order  to  insure  equilibrium?  The 
weights  are  50,  73.4,  16.93,  and  217  lbs. 

Answer:  Invert  slide,  set  5B  to  380D  and  under  the  number  of  pounds 
in  B  find  the  answer  in  feet  on  D  as  follows:    3.80,  2.59,  11.22,  .875. 

(6).  Find  the  supporting  forces  on  the  two  ends  of  a  beam  22.3  feet 
long  under  the  action  of  a  movable  load  of  625  pounds.  Reactions  are  desired 
when  the  load  is  concentrated  at  distances  of  2,  4,  6,  8, 10,  12  feet  from  the  left- 
hand  end  of  the  beam.  Answers  to  be  given  to  the  nearest  pound.  Answers 
55-570,  112-513,  168-457,  224-401,  etc. 


—  32 


(7).  Find  the  number  of  amperes  required  to  run  a  motor  developing 
15  horsepower  at  voltages  of  110,  220,  550,  and  750;  using  the  formula  H.  P. — 

VXA 

.     Answer:     102,  51,  20.4,  14.9. 

746 

(8).  At  a  pressure  of  two  atmospheres  a  certain  quantity  of  a  eartain 
gas  has  a  volume  of  22  cubic  feet.  Assuming  Boyle's  law  to  hold,  that  is; 
PRESSURE  X  VOLUME  =  CONSTANT,  find  the  volume  of  the  gas  at 
pressures  of  33,  37,  40,  45,  46  pounds  per  square  inch. 

Answers:     19.6,  17.5,  16.2,  14.4,  14.1. 

(9).  If  a  pump  A  of  capacity  47  cubic  feet  per  minute  can  empty  an 
excavation  in  12  l/i  days  at  a  cost  of  $3.45  per  day,  and  if  a  pumpB,  capacity 
38  cubic  feet  per  minute,  which  runs  at  a  cost  of  $2.83,  is  also  available,  which 
will  be  the  cheaper  to  use  ?     Answer:     A. 

(10).     State  the  simplest  setting  for  (9). 

Answer:  Invert  slide.  Set  47B  to  1225D,  R  to  38B,  283B  to  R,  R  to 
1225D.     If  the  number  on  B  under  R  is  greater  than  345,  A  is  the  cheaper. 

THE  FORMS  XY  =  B  AND  X2Y  =  B. 

(1).  Find  the  dimensions  of  a  rectangle,  its  sides  being  determined  to 
the  nearest  inch  only,  so  that  the  rectangle  shall  contain  as  nearly  as  possible 
1317  square  inches.     The  short  side  of  the  rectangle  must  be  over  25  in  long. 

Answer:    28  X  47  in. 

(2).  Determine  the  dimensions  of  a  right  triangle  so  it  will  contain,  at 
least,  and  as  near  317  square  inches  as  possible.  The  sides  are  to  be  measured 
to  the  nearest  J^  in.     The  short  leg  of  the  triangle  must  be  15  in.  or  over. 

Answer:    29  J^  X  21 J^  in. 

(3).  A  sewer  must  have  a  cross  sectional  area  of  at  least  716  square 
inches  and  is  elliptical  in  section.  Find  the  lengths  of  the  major  and  minor 
axes  to  the  nearest  Inch.  The  minor  axis  must  be  between  20  and  25  inches 
long.    Area  of  ellipse  is  .7854  a  X  6,  where  a  is  the  major  and  b  the  minor 

axis.    Answer:    24  X  38. 

(4).  Determine  the  number,  width  and  thickness  of  steel  plates  for 
the  flange  of  a  girder,  so  that  the  gross  cross-sectional  area  of  the  plates  shall 
exceed  by  an  amount  as  small  as  possible,  23.7  square  inches.  The  widths  of 
the  plates  can  vary  by  half  inches  from  10  to  12  inches,  and  in  thickness  from 
%  to  Ys  by  sixteenths.  The  plates  must  all  be  the  same  width  and  should  vary 
in  thickness  as  little  as  possible. 

Answer:    3  PI.  1VA  X  V%  in.,  1  PL  11 K  X  Vie  in. 

(5).  Determine  plates  required  under  conditions  which  are  identical 
with  problem  (4),  except  that  the  gross  area  must  be  43.6  and  the  plates  may 
vary  in  width  by  half  inches  from  14  to  18  in. 

Answer:     5  PL  14  X  Ys  in.,  or  4  PL  17 ^  X  Ys  in. 


—  33  — 


(6).     Design  the  section  of  a  rectangular  wooden  beam  using  the 

fbh* 

formula  M  = ,  where  M  =  31250  inch  pounds,  and  /  =  980  lbs.  per 

6 
square  inch.     The  minimum  value  of  6  is  two  inches.     Both  b  and  h  must  be 
multiples  of  two.    Answer:    2  X  10  in. 

(7).  Design  a  beam  under  the  same  conditions  as  problem  (6)  where 
M  =  72500  inch  pounds  and  /  =  1000  lbs.  per  square  inch.  Minimum  value 
for  6  is  6  in.    Answer:    6  X  10  in. 

(8).  Determine  b  and  h  for  a  rectangular  reinforced  concrete  beam 
where  M  =  125  b  h2.  M  =  1500000  inch  pounds,  h  must  be  at  least  double 
b,  and  b  at  least  13  in.  Determine  the  dimensions  to  the  nearest  J^  in.  Answer: 
13^x30  in. 

CUBES. 

(1).     Cube  the  following  numbers;    728,  8.07,  55.9,  10.17. 
Answers:    386000000,  526,  174700,  1052. 

(2).  Cube  the  following  with  the  slide  inverted:  9.77,  100.5,  .246, 
.0133.    Answers:    933,  1015000,  .01489,  .000002353. 

(3).  Solve  the  following  with  the  slide  direct:  (15.1)?,  (2.34)1,  (.0327)1. 
(21/3i)I.    Answers:    58.7,  3.58,  .00591,  17/32  or  .532. 

(4).  Deduce  the  setting  for  xf  with  the  slide  inverted,  when  x  lies 
between  1  and  10.  Answer:  xRC  to  xD,  under  IRC  read  answer  on  D,  or 
xB  to  xLA,  under  IRC  read  answer  on  D. 

(5).  With  the  scale  inverted  find  the  following;  expressing  the  answer 
to  the  nearest  thirty-second  of  an  inch.     (3.71)1,  (7A6)I,  (1V8)I,  (47.5)i. 

229    9    38  10480 

Answers      ,  — ,  — , . 

32    32  32      32 


CUBE  ROOTS. 

(1).    Find  the  cube  roots  of  the  following  with  the  slide  direct:    2.613, 
17.8  .00321,  1793,  .0488.    Answers:     1.377,  2.61,  .1475,  12.15,  .3655. 

(2).     Find  the  cube  roots  of  the  following  with  the  slide  inverted; 
3.93,  14.16,  .000444,  .01331,  571.    Answers:     1.578,  2.42,  .0763,  .237,  8.30. 


(3).     Solve  the  following:  f/{8.93)2,    f/(71.7)2    ^(IA2S)\   f/(.0321)K 
Answers:    4.30,  17.26,  1.265,  .1010. 

(4).  Deduce  the  general  setting  for  the  type  Hi,  with  the  slide  inverted. 
Answer:  1C  to  HD.  Look  for  coincidences  between  A  and  B,  or  C  and  D. 
The  numbers  coinciding  will  give  the  answer. 

(5).     Solve  the  equation  x|  =  18.17.    Answer:    x  =  77.5. 

(6).     Deduce  the  setting  for  the  solution  of  the  type  shown  in  (5). 
Answer:     Set  right-hand  member  on  A  to  right-hand  member  on  B  and 
read  x  on  D  under  1C.     Slide  inverted. 


—  34 


(7).     Solve  the  equation  x%  =  9.81.     Answer:    x  =  4.58. 

(8).  Deduce  the  setting  for  the  solution  of  the  type  shown  in  (7)  with 
the  slide  direct.  Answer:  R  to  right-hand  member  on  D,  move  the  slide  until 
the  same  number  on  B  is  under  R,  that  appears  on  D  under  1C.  This  number 
will  be  the  required  value  of  x. 

(9).  A  right  circular  cone  contains  347  cubic  inches  and  the  radius 
of  the  base  is  7{  in.  What  will  be  the  volume  of  a  similar  cone,  the  radius  of 
the  base  being  8.19  in.  ?     Answer:     500  cu.  in. 

(10).     If  a  sphere  of  (a)  inches  radius  contains  400  gallons,  what  will  be 
the  radius  of  a  sphere  containing  600  gallons.     Answer:     1.144  a. 

(11).     Find  in  problem  (10)  the  radii  of  the  two  spheres  in  feet. 
Answer:    2.33  ft.  and  2.67  ft. 


LOGARITHMS,  INVOLUTION  AND  EVOLUTION. 

(1).     Find  the  logarithms  of  the  following  numbers  to  three  places 
in  the  mantissa:     2181,  131.7,  .983,  .0436,  7.94. 

Ansivers:    3.339,  2.120,  9.993-10,  8.639-10,  0.900. 

(2).     Find  the   numbers   corresponding  to  the  following  logarithms; 
2.736,  7.1316-10,  .4717,  6.979-10.     Answers:     544.5,  .001354,  .2963,  .000953. 

(3).     Find  the  fifth  power  of  3.72  by  two  distinct  methods. 
Answer:     712. 

(4).     Solve  (1.217)1  by  the  use  of  logarithms.     Answer:     1.387. 

(5).     Solve  (8.44)127  and  (.00511)311.     Answers:     15.01  and  .0000000747. 

(6).    Solve  (.0132)-"05  and  -31\/1729.     Answers:     .173  and  27.8  billions. 

EQUATIONS  INVOLVING  FRACTIONAL  POWERS. 


wx 
We  f 

(1).    Solve  the  equation  S  = ,  when  W  is  10  tons,  x 

f 


100  feet, 


w  =  .27  pounds  per  cubic  inch,  /  =  10000  lbs.  per  square  inch.     Value  for  e 

will  be  found  on  the  reverse  of  the  rule.     Express  the  answer  in  square  inches. 

Answer:     2.066. 

99318  X  D3"55 

(2).    Using  Hodgkinson's  formula  W  =  ,  find  W  when  D 

L17 

=  2.5  andL  =  17.8.     Answer:     19230  lbs.  per  sq.  in. 

(3).     Solve  the  equation  x218  =  3.43.     Answer:    x  =  1.760. 

14  X  3.1416  X  (5)J 


(4).     Solve  for  t:     t  = 


15     

—V  64.4  X  12 
4 


Answer:    t  =  23.6. 


(5).     Solve  for  q  in  the  equation:     q  =  3.01  X  16.7  X  (2.32)1.53 
Answer:  182.2 


—  35 


*S^> 


TRIGONOMETRIC  COMPUTATIONS. 

(1).     Find  sin  32°  14',  sin  85°  38',  tan  19°  14',  tan  22°  41'. 
Answers:     .533,  .997,  .349,  .418. 

(2).     Find  cos  29°  17',  cos  73°  04',  cot  89°  19',  cot  12°  51'. 
Answers:     .872,  .291,  .01193,  4.38. 

(3).    Find  sec  45°  09',  sec  73°  33',  esc  11°  13',  esc  36°  28'. 
Answers:     1.419,  3.53,  5.14,  1.685. 

(4).     Find  sin  00°  18',  tan  00°  07',  sin  00°  41',  tan  01°  03'. 
Answers:     .00524,  .00204,  .01193,  .01833. 

(5).     Perform  the  following  operations: 

sin  41°  03'  X  386  tan  20°  10'  X  31.3 
36.1  X  sin  30°  35', 


sin  10°  17'  tan  43°  16' 

Answers:    18.4,  1420,  12.21. 

(6).    Find  log  sin  39°  19',  log  sin  15°  52',  log  tan  40°  40',  log  cot  49°  35', 
log  sec  86°  17',  log  esc  4°  44'. 

Answers:     9.802-10,  9.437-10,  9.934-10,  9.930-10, 1.188, 1.083. 
(7).    Solve  the  triangle  as  given. 


31.3 


104.6 

Answers:    A  =  16°39',B  =  73°  21',  c  =  109.2. 
(8).    Solve  the  triangle  as  given. 


65°40' 


Answers:    A  =  24°  20',  b  =  108.5,  a  =  49.1. 
(9).    Solve  the  triangle  as  given. 


30°03: 


Answers:     A  =  22°  14',  B  =  127°  43',  b  =  12.86. 


—  36  — 


(10.     Solve  the  triangle  as  given. 
17°  20' 


12°  13' 


Answers;    B  =  150°  27',  b  =  7.76,  a  =  4.70. 


VARIATION  AS  THE  SQUARE. 

(1).  A  circle  has  an  area  of  34.71  square  inches.  What  will  be  the 
diameter  of  a  circle  whose  area  is  3  34  times  as  large?     Answer:     12  in. 

(2).  An  equilateral  triangle  has  its  sides  3.17  in.  long  and  has  an 
area  of  4.33  sq.  in.  What  is  the  area  of  an  equilateral  triangle  whose  side  is 
3.47  in.;  one  whose  side  is  7.62;  one  whose  side  is  17.1  ? 

Answers:    5.18,  25.01, 125.9  in. 

(3).  A  bar  of  steel  1  ft.  long  and  \\i  in.  in  diameter  weighs  4.17  lbs. 
What  will  be  the  weight  of  bars  1  ft.  long  and  with  the  following  diameters: 
134  1%  and  2  ^  in.  ?    Answers:  3.38,  8.18, 16.6  lbs. 

(4).  In  problem  (1)  if  e  is  the  area  of  the  given  circle,  6  the  ratio  of 
the  area  of  the  required  to  the  given  circle,  find  the  general  setting  for  problems 
of  this  type.  Answer:  Set  1RB  to  eRA,  R  to  bLB,  79C  to  70D,  answer  under 
runner  on  C. 

(5).  A  beam  28  feet  long  is  loaded  with  a  uniform  load  such  that  the 
moment  at  the  center  is  7070000  inch  pounds.  Find  the  moments  at  2  foot 
intervals  over  the  beam. 

Answers  in  tens  of  thousands  of  inch  pounds. 

0  2  4  6  8  10  12  14  16   etc. 

0        188        346        476         577        649        693         707         693 

(6).  With  the  parabola  as  given  with  the  vertex  at  (a)  find  the  distance 
(b)  so  that  c  is  21.5  feet.     Answer:    61.4  ft. 


34.5 


< lOO 


> 

-  .       • 

—  37  —  . 


(7).     A  parabola  as  given  with  the  vertex  at  a;  dc  =  238',  ad  =  440'. 
Determine  perpendicular  offsets  to  the  parabola  from  points  40  feet  apart 


on  ad    Answer: 

a        40    80 

120 

160 

200 

240 

280 

320 

360 

400 

0    1.9   7.9 

17.7 

31.5 

49.3 

71.0 

96.8 

126.0 

160 

197 

SPECIAL  FORMS. 


(1 
(2 
(3 
(4 
(5 

(6 

(7 

(8 
(9 

(10 


(8.72)2 

Compute .     Answer:     .289. 

32.2  X  8.17 


(91.5)2 
Compute  A  /—  — .    Answer:    3.90. 


4 


32.2  X   17.17 


.925  X   (13.11)2 

Compute .     Answer:    2.47. 

64.4 


\/MA  X  >/73.8 

Compute .     Answer:     15.81. 

4.36 


13.16  X  \/.036 

Compute .     Answer:     .1290. 

19.37 


12.718  X  3.4 

Compute  8.88  A  .     Answer:     3.14. 

73.7 


4 


\/(3.13)3 

Compute .    Answer:     .420. 

13.19 

Compute  8.9  (.032)f.     Answer:     .0510. 


\/3.18  X  9.72 

Compute .     Answer:     13.02. 

1.331 


132.9  X  (1.313)2 

Compute  .»  .     Answer:     7.71. 

(.977)2 


4 


—  38 


TABLE  OF  CONTENTS. 

Page 
Preface 2 

Accuracy  and  Significant  Figures .        .  3 

Description  of  the  Rule 5 

Decimal  Point  and  Reading  the  Scale 6 

Polyphase  Slide  Rule 14 

Principle  of  the  Rule .6 

Equivalent  Ratios 7 

Squares  and  Square  Roots             .                       9 

Multiplication  and  Division 10 

Inverted  Slide 13 

Forms  xy  =  B,  x2y  =  B            .               15 

Cubes 17 

Cube  Roots 18 

Logarithms .        .  20 

Involution  and  Evolution  (General) 20 

Equations  Involving  Fractional  Exponents                               ....  21 

Trigonometric  Computations 22 

Variation  as  the  Square 25 

Special  Forms .            26 

General  Conditions 28 

Problems . 29 


»    •      * 

c       ■  ■ 


SLIDE    RULES    OF    ALL    KINDS. 

We  are  the  largest  manufacturers  of  slide  rules  in  America  and  have  the 
largest  assortment.    Some  of  the  best  known  of  our  slide  rules  are  shown  below. 

MANNHEIM 


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T^r,^,fV^rl,.,,.,,,,l1.l|{.,.1^rp(l. Mj ..j-^.i 


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Mannheim  Slide  Rules  can  be  used  to  advantage  in  every  line  of  busi- 
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the  slide. 

POLYPHASE 


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hrimffiwii'rtii'tfifiij       89* 

j  _  ,  i  j  i  I  ,py 


The  Polyphase  Slide  Rule  is  of  the  Mannheim  type,  but  has  an  inverted 
C  scale  and  a  scale  of  cubes  in  addition  to  the  regular  Mannheim  scales.  This 
arrangement  facilitates  the  solution  of  many  problems  involving  three  factors, 
as  well  as  many  powers  and  roots. 

DUPLEX 

Duplex  Slide  Rules  are  practically  two  rules  in  one,  with  all  the  scales 
on  the  exterior  faces,  where  they  may  be  readily  seen  and  utilized  by  merely 
turning  the  rule  over. 

POLYPHASE    DUPLEX 


a-t^—a' 


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The  Polyphase  Duplex  Slide  Rules  are  a  combination  of  the  Polyphase 
and  the  Duplex  Rules,  with  the  addition  of  several  special  scales. 

K  &  E    STADIA    SLIDE    RULE 


10  80  JO  *-  ^o  tip 

liiiilinil,inN«*Ji..,!iuH.,»i.!ilii.'ilililihlll,liiiliMllW,Wili,'d',ai|ii,il,iii]„il,  ■!.    ,!■■  J  -.1    ., .  iV. UimlnJutoUil 


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This  form  of  Stadia  Slide  Rule  is  remarkable  for  its  simplicity.  When 
the  stadia  rod  reading  and  the  angle  of  elevation  or  depression  of  the  telescope 
are  known,  the  horizontal  distance  and  vertical  height  can  be  obtained  at 
once  by  one  setting  of  slide. 


Write  for  our  Slide  Rule  Booklet,  fully  describing  the  foregoing,  as  well 
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